Step 1: Identifying the Mathematical Framework

To evaluate the definite integral $I = \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$, we employ the Fundamental Theorem of Calculus (Part II). This theorem states that if $F(x)$ is the antiderivative of $f(x)$, then $\int_{a}^{b} f(x) \, dx = F(b) - F(a)$. [Fundamental Theorem of Calculus].

Step 2: Finding the Antiderivative

We identify the integrand $f(x) = \cos 2x$. To find the antiderivative $F(x)$, we apply the method of integration by substitution or the reverse chain rule for trigonometric functions. Since $\frac{d}{dx}(\sin(kx)) = k \cos(kx)$, it follows that $\int \cos(kx) \, dx = \frac{1}{k} \sin(kx) + C$.

Setting $k = 2$: $F(x) = \int \cos 2x \, dx = \frac{\sin 2x}{2}$

Step 3: Applying Limits of Integration

Substituting the upper limit $b = \frac{\pi}{2}$ and the lower limit $a = 0$ into the antiderivative:

$I = \left[ \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}$

$I = \frac{1}{2} \left[ \sin\left(2 \cdot \frac{\pi}{2}\right) - \sin(2 \cdot 0) \right]$

Step 4: Algebraic Evaluation

Evaluate the trigonometric terms:

• $\sin(2 \cdot \frac{\pi}{2}) = \sin(\pi) = 0$ [Trigonometric Identity]
• $\sin(0) = 0$

$I = \frac{1}{2} [0 - 0] = 0$

Visual Representation of the Integrand

The integral represents the net signed area between the curve $y = \cos 2x$ and the $x$-axis from $0$ to $\frac{\pi}{2}$. Because the area from $0$ to $\frac{\pi}{4}$ (positive) exactly cancels the area from $\frac{\pi}{4}$ to $\frac{\pi}{2}$ (negative), the result is zero.

Final Solution: $\int_{0}^{\frac{\pi}{2}} \cos 2x \, dx = 0$

Step 1: Decomposition of the Definite Integral

Given the definite integral: $I = \int_{0}^{\frac{\pi}{4}} (2\sec^2 x + x^3 + 2) \, dx$.
By the Linearity Property of Integration [$\int_{a}^{b} [f(x) + g(x)] dx = \int_{a}^{b} f(x) dx + \int_{a}^{b} g(x) dx$], we decompose the expression into three distinct integrable terms:

$I = 2 \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx + \int_{0}^{\frac{\pi}{4}} x^3 \, dx + 2 \int_{0}^{\frac{\pi}{4}} 1 \, dx$

Step 2: Identifying Fundamental Antiderivatives

We apply the Fundamental Theorem of Calculus Part II: $\int_{a}^{b} f(x) dx = F(b) - F(a)$, where $F'(x) = f(x)$.

• For $\int \sec^2 x \, dx$, the antiderivative is $\tan x$ [Standard Trigonometric Integral].
• For $\int x^3 \, dx$, the antiderivative is $\frac{x^4}{4}$ [Power Rule: $\int x^n dx = \frac{x^{n+1}}{n+1}$].
• For $\int 1 \, dx$, the antiderivative is $x$.

Step 3: Evaluating the Antiderivatives

Applying the limits $[0, \frac{\pi}{4}]$ to each term:

$I = 2[\tan x]_{0}^{\frac{\pi}{4}} + \left[ \frac{x^4}{4} \right]_{0}^{\frac{\pi}{4}} + 2[x]_{0}^{\frac{\pi}{4}}$

Calculating the values:

1. Term 1: $2(\tan(\frac{\pi}{4}) - \tan(0)) = 2(1 - 0) = 2$
2. Term 2: $\left( \frac{(\frac{\pi}{4})^4}{4} - 0 \right) = \frac{\pi^4}{4^4 \cdot 4} = \frac{\pi^4}{256 \cdot 4} = \frac{\pi^4}{1024}$
3. Term 3: $2(\frac{\pi}{4} - 0) = \frac{2\pi}{4} = \frac{\pi}{2}$

Step 4: Algebraic Summation

Summing the results obtained in Step 3:

$I = 2 + \frac{\pi^4}{1024} + \frac{\pi}{2}$

Visual Representation of Area under the curve $f(x) = 2\sec^2 x + x^3 + 2$

Final Solution: $2 + \frac{\pi}{2} + \frac{\pi^4}{1024}$

Step 1: Theoretical Foundation

To evaluate the definite integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x \, dx$, we utilize the standard integral formula for the cosecant function. The antiderivative of $\csc x$ is given by:

$\int \csc x \, dx = \ln |\csc x - \cot x| + C$ [Standard Integral Calculus Identity]

Step 2: Applying the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, Part II, if $F(x)$ is the antiderivative of $f(x)$, then:

$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$

Substituting our limits $a = \frac{\pi}{6}$ and $b = \frac{\pi}{4}$:

$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x \, dx = \left[ \ln |\csc x - \cot x| \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}$

Step 3: Evaluating at the Boundaries

Substitute the upper limit $x = \frac{\pi}{4}$:

$F(\frac{\pi}{4}) = \ln |\csc(\frac{\pi}{4}) - \cot(\frac{\pi}{4})| = \ln |\sqrt{2} - 1|$

[Using trigonometric ratios: $\csc(\frac{\pi}{4}) = \sqrt{2}$, $\cot(\frac{\pi}{4}) = 1$]

Substitute the lower limit $x = \frac{\pi}{6}$:

$F(\frac{\pi}{6}) = \ln |\csc(\frac{\pi}{6}) - \cot(\frac{\pi}{6})| = \ln |2 - \sqrt{3}|$

[Using trigonometric ratios: $\csc(\frac{\pi}{6}) = 2$, $\cot(\frac{\pi}{6}) = \sqrt{3}$]

Step 4: Final Algebraic Simplification

Subtracting the values:

Result $= \ln |\sqrt{2} - 1| - \ln |2 - \sqrt{3}|$

Using the logarithmic quotient rule $\ln(a) - \ln(b) = \ln(\frac{a}{b})$:

Result $= \ln \left| \frac{\sqrt{2} - 1}{2 - \sqrt{3}} \right|$

Rationalize the denominator to simplify the expression:

$\frac{\sqrt{2} - 1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2\sqrt{2} + \sqrt{6} - 2 - \sqrt{3}}{4 - 3} = 2\sqrt{2} + \sqrt{6} - 2 - \sqrt{3}$

Visual Representation of the Integration Area

Final Solution: \ln \left| \frac{\sqrt{2} - 1}{2 - \sqrt{3}} \right| \text{ or } \ln(2\sqrt{2} + \sqrt{6} - 2 - \sqrt{3})

Step 1: Application of Trigonometric Identities

To evaluate the integral $I = \int_{0}^{\pi} (\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}) \, dx$, we first simplify the integrand using the double-angle identity for cosine: $\cos(2\theta) = \cos^2 \theta - \sin^2 \theta$.

By rearranging the given integrand:

$\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})$

Applying the identity with $\theta = \frac{x}{2}$:

$-(\cos(2 \cdot \frac{x}{2})) = -\cos(x)$

[By Trigonometric Identity: $\cos(2A) = \cos^2 A - \sin^2 A$]

Step 2: Reformulation of the Integral

Substitute the simplified expression back into the definite integral:

$I = \int_{0}^{\pi} -\cos(x) \, dx$

$I = -\int_{0}^{\pi} \cos(x) \, dx$

Step 3: Finding the Antiderivative

The integral of $\cos(x)$ with respect to $x$ is $\sin(x)$. [Per the Fundamental Theorem of Calculus Part II, if $F'(x) = f(x)$, then $\int_{a}^{b} f(x)dx = F(b) - F(a)$].

$I = -[\sin(x)]_{0}^{\pi}$

Step 4: Evaluating at Boundaries

Apply the limits of integration ($0$ and $\pi$):

$I = -(\sin(\pi) - \sin(0))$

Evaluating the trigonometric values:

$\sin(\pi) = 0$

$\sin(0) = 0$

$I = -(0 - 0) = 0$

Visual Representation of the Function

Final Solution: $0$

Step 1: Theoretical Foundation

The integral of the trigonometric function $\tan x$ is a fundamental antiderivative derived from the logarithmic form of the natural logarithm of the secant function. [Per the Fundamental Theorem of Calculus, $\int_{a}^{b} f(x) dx = F(b) - F(a)$ where $F'(x) = f(x)$].

Step 2: Integration of the Indefinite Form

The indefinite integral is defined as:

$\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx$

By applying the substitution method, let $u = \cos x$, then $du = -\sin x \, dx$, which implies $\sin x \, dx = -du$.

$\int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos x| + C = \ln|\sec x| + C$ [Using the logarithmic property $-\ln(a) = \ln(a^{-1})$].

Step 3: Application of Definite Limits

Applying the limits $[0, \frac{\pi}{4}]$ to the antiderivative $\ln|\sec x|$:

$I = \left[ \ln|\sec x| \right]_{0}^{\frac{\pi}{4}}$

$I = \ln|\sec(\frac{\pi}{4})| - \ln|\sec(0)|$

Step 4: Evaluation of Trigonometric Values

Given the standard trigonometric ratios:

• $\sec(\frac{\pi}{4}) = \frac{1}{\cos(\frac{\pi}{4})} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$
• $\sec(0) = \frac{1}{\cos(0)} = \frac{1}{1} = 1$

Step 5: Final Calculation

$I = \ln|\sqrt{2}| - \ln|1|$

Since $\ln(1) = 0$ [Per the logarithmic identity $\ln(1) = 0$]:

$I = \ln(2^{1/2}) - 0$

$I = \frac{1}{2} \ln(2)$

Geometric Visualization

Final Solution: $\frac{1}{2} \ln 2$

Step 1: Decomposition of the Integral

The given definite integral is:

$I = \int_{0}^{2} \frac{6x+3}{x^2+4} dx$

By the additive property of integrals [Linearity of the integral operator], we can split the integrand into two distinct parts to simplify the integration process:

$I = \int_{0}^{2} \frac{6x}{x^2+4} dx + \int_{0}^{2} \frac{3}{x^2+4} dx$

Step 2: Evaluating the First Integral ($I_1$)

Let $I_1 = \int_{0}^{2} \frac{6x}{x^2+4} dx$.

Use the method of $u$-substitution. Let $u = x^2 + 4$.

Then, the derivative is $\frac{du}{dx} = 2x$, which implies $du = 2x dx$.

Since the numerator is $6x dx = 3(2x dx) = 3 du$, we perform the change of limits:

• When $x = 0$, $u = 0^2 + 4 = 4$.
• When $x = 2$, $u = 2^2 + 4 = 8$.

$I_1 = \int_{4}^{8} \frac{3}{u} du = 3 [\ln|u|]_{4}^{8}$

$I_1 = 3 (\ln 8 - \ln 4) = 3 \ln\left(\frac{8}{4}\right) = 3 \ln 2$

Step 3: Evaluating the Second Integral ($I_2$)

Let $I_2 = \int_{0}^{2} \frac{3}{x^2+4} dx$.

We use the standard integral formula: $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$ [Integration of rational functions].

Here, $a^2 = 4$, so $a = 2$.

$I_2 = 3 \int_{0}^{2} \frac{1}{x^2 + 2^2} dx = 3 \left[ \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

$I_2 = \frac{3}{2} \left[ \tan^{-1}\left(\frac{2}{2}\right) - \tan^{-1}\left(\frac{0}{2}\right) \right]$

$I_2 = \frac{3}{2} (\tan^{-1}(1) - \tan^{-1}(0)) = \frac{3}{2} \left( \frac{\pi}{4} - 0 \right) = \frac{3\pi}{8}$

Step 4: Combining Results

The total integral is the sum of $I_1$ and $I_2$:

$I = 3 \ln 2 + \frac{3\pi}{8}$

Visualization of the Area under the Curve $f(x) = \frac{6x+3}{x^2+4}$

Final Solution: $3 \ln 2 + \frac{3\pi}{8}$

Step 1: Decomposition of the Integral

By the linearity property of definite integrals, we can split the given integral into two distinct parts [Per the Linearity Property of Integrals]:

$\int_{0}^{1} (x e^x + \sin \frac{\pi x}{4}) \, dx = \int_{0}^{1} x e^x \, dx + \int_{0}^{1} \sin \frac{\pi x}{4} \, dx$

Step 2: Integration of the First Term ($I_1 = \int_{0}^{1} x e^x \, dx$)

To evaluate this integral, we employ the Integration by Parts formula: $\int u \, dv = uv - \int v \, du$ [Per the Integration by Parts Theorem].

• Let $u = x \implies du = dx$
• Let $dv = e^x \, dx \implies v = e^x$

Applying the formula:

$I_1 = [x e^x]_{0}^{1} - \int_{0}^{1} e^x \, dx$

$I_1 = (1 \cdot e^1 - 0 \cdot e^0) - [e^x]_{0}^{1}$

$I_1 = e - (e^1 - e^0) = e - e + 1 = 1$

Step 3: Integration of the Second Term ($I_2 = \int_{0}^{1} \sin \frac{\pi x}{4} \, dx$)

We use the standard integral formula $\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C$ [Per the Integration of Trigonometric Functions].

$I_2 = \left[ -\frac{1}{\pi/4} \cos \frac{\pi x}{4} \right]_{0}^{1}$

$I_2 = -\frac{4}{\pi} \left[ \cos \frac{\pi(1)}{4} - \cos \frac{\pi(0)}{4} \right]$

$I_2 = -\frac{4}{\pi} \left( \cos \frac{\pi}{4} - \cos 0 \right)$

Since $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\cos 0 = 1$:

$I_2 = -\frac{4}{\pi} \left( \frac{1}{\sqrt{2}} - 1 \right) = \frac{4}{\pi} \left( 1 - \frac{1}{\sqrt{2}} \right)$

Step 4: Summation of the Components

Combining $I_1$ and $I_2$:

Result = $1 + \frac{4}{\pi} \left( 1 - \frac{1}{\sqrt{2}} \right)$

Simplifying the expression within the parenthesis:

Result = $1 + \frac{4}{\pi} - \frac{4}{\pi\sqrt{2}} = 1 + \frac{4}{\pi} - \frac{2\sqrt{2}}{\pi}$

Final Solution: $1 + \frac{4 - 2\sqrt{2}}{\pi}$

Step 1: Definition of the Definite Integral as a Limit of a Sum

According to the Fundamental Theorem of Calculus, the definite integral $\int_{a}^{b} f(x) dx$ can be evaluated using the limit of a Riemann sum as $n \to \infty$:

$\int_{a}^{b} f(x) dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh)$

Where:

• $a = -1$
• $b = 1$
• $h = \frac{b-a}{n} = \frac{1 - (-1)}{n} = \frac{2}{n}$

Step 2: Substitution of Given Function and Variables

Given $f(x) = x + 1$. We evaluate $f(a + rh)$:

$f(-1 + rh) = (-1 + rh) + 1 = rh$

Substituting this into the summation formula:

$\int_{-1}^{1} (x+1) dx = \lim_{n \to \infty} \left( \frac{2}{n} \right) \sum_{r=0}^{n-1} \left( r \cdot \frac{2}{n} \right)$

Step 3: Algebraic Simplification of the Summation

Factoring out constants from the summation [Per the Summation Property $\sum c \cdot a_r = c \sum a_r$]:

$= \lim_{n \to \infty} \left( \frac{2}{n} \right) \left( \frac{2}{n} \right) \sum_{r=0}^{n-1} r$

Using the identity for the sum of the first $(n-1)$ natural numbers, $\sum_{r=0}^{n-1} r = \frac{(n-1)n}{2}$ [By Arithmetic Progression Sum formula]:

$= \lim_{n \to \infty} \frac{4}{n^2} \cdot \frac{(n-1)n}{2}$

Step 4: Evaluating the Limit

Simplifying the expression within the limit:

$= \lim_{n \to \infty} \frac{4n(n-1)}{2n^2} = \lim_{n \to \infty} 2 \left( \frac{n^2 - n}{n^2} \right)$

$= \lim_{n \to \infty} 2 \left( 1 - \frac{1}{n} \right)$

As $n \to \infty$, $\frac{1}{n} \to 0$. Thus:

$= 2(1 - 0) = 2$

Step 5: Geometric Visualization

The integral represents the area of a right-angled triangle with base $b = 2$ and height $h = 2$. Area $= \frac{1}{2} \times 2 \times 2 = 2$.

Final Solution: 2

Step 1: Decomposition of the Integrand

The integral to be evaluated is $I = \int_{0}^{1} \frac{2x+3}{5x^2+1} dx$. We begin by splitting the integral into two distinct parts to handle the linear term and the constant term separately [By the Additivity Property of Integrals]:

$I = \int_{0}^{1} \frac{2x}{5x^2+1} dx + \int_{0}^{1} \frac{3}{5x^2+1} dx$

Step 2: Evaluating the First Integral ($I_1$)

Let $I_1 = \int_{0}^{1} \frac{2x}{5x^2+1} dx$. We apply the method of substitution [By Integration by Substitution]:

Let $u = 5x^2 + 1$. Then, the derivative is $\frac{du}{dx} = 10x$, which implies $2x dx = \frac{1}{5} du$.

Adjusting the limits: When $x=0$, $u=1$. When $x=1$, $u=6$.

$I_1 = \frac{1}{5} \int_{1}^{6} \frac{1}{u} du = \frac{1}{5} [\ln|u|]_{1}^{6}$

$I_1 = \frac{1}{5} (\ln 6 - \ln 1) = \frac{\ln 6}{5}$

Step 3: Evaluating the Second Integral ($I_2$)

Let $I_2 = \int_{0}^{1} \frac{3}{5x^2+1} dx$. We factor out the constant 5 from the denominator to match the standard form $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a})$:

$I_2 = 3 \int_{0}^{1} \frac{1}{5(x^2 + \frac{1}{5})} dx = \frac{3}{5} \int_{0}^{1} \frac{1}{x^2 + (\frac{1}{\sqrt{5}})^2} dx$

Using the standard integral formula [By the Integral of $1/(x^2+a^2)$]:

$I_2 = \frac{3}{5} \left[ \frac{1}{1/\sqrt{5}} \tan^{-1} \left( \frac{x}{1/\sqrt{5}} \right) \right]_{0}^{1}$

$I_2 = \frac{3\sqrt{5}}{5} \left[ \tan^{-1}(\sqrt{5}x) \right]_{0}^{1} = \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$

Step 4: Combining Results

Summing the two components $I = I_1 + I_2$:

$I = \frac{\ln 6}{5} + \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$

Visual Representation of the Integrand

Final Solution: $\frac{1}{5} \ln 6 + \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$

Step 1: Identification of the Integral

The given definite integral is defined as:

$I = \int_{0}^{\frac{\pi}{4}} \sin(2x) \, dx$

Step 2: Determining the Antiderivative

To evaluate the integral, we identify the antiderivative of the function $f(x) = \sin(2x)$. [Per the Fundamental Theorem of Calculus Part II].

We use the substitution method or the direct formula for linear arguments: $\int \sin(ax) \, dx = -\frac{\cos(ax)}{a} + C$.

Here, $a = 2$, therefore:

$\int \sin(2x) \, dx = -\frac{\cos(2x)}{2}$

Step 3: Application of the Fundamental Theorem of Calculus

Applying the limits of integration from $0$ to $\frac{\pi}{4}$:

$I = \left[ -\frac{\cos(2x)}{2} \right]_{0}^{\frac{\pi}{4}}$

$I = -\frac{1}{2} \left[ \cos(2x) \right]_{0}^{\frac{\pi}{4}}$

Step 4: Evaluating at Boundaries

Substitute the upper and lower limits into the expression:

$I = -\frac{1}{2} \left[ \cos\left(2 \cdot \frac{\pi}{4}\right) - \cos(2 \cdot 0) \right]$

$I = -\frac{1}{2} \left[ \cos\left(\frac{\pi}{2}\right) - \cos(0) \right]$

Step 5: Trigonometric Simplification

Recall the standard trigonometric values: $\cos\left(\frac{\pi}{2}\right) = 0$ and $\cos(0) = 1$. [By definition of unit circle coordinates].

$I = -\frac{1}{2} \left[ 0 - 1 \right]$

$I = -\frac{1}{2} \left[ -1 \right]$

$I = \frac{1}{2}$

Step 6: Visual Representation

Final Solution: \frac{1}{2}

Step 1: Establishing the Integral and Substitution Method

Given the definite integral:

$$I = \int_{2}^{3} \frac{x}{x^2+1} dx$$

To evaluate this, we identify a function $f(x) = x^2 + 1$ such that its derivative $f'(x) = 2x$ is proportional to the numerator. We utilize the method of integration by substitution. [Per the Integration by Substitution Rule].

Step 2: Performing Variable Substitution

Let $u = x^2 + 1$. Then, differentiating both sides with respect to $x$ yields:

$$\frac{du}{dx} = 2x \implies du = 2x dx \implies x dx = \frac{1}{2} du$$

Step 3: Adjusting Limits of Integration

Since the variable of integration changes from $x$ to $u$, we must transform the limits of integration accordingly [Fundamental Theorem of Calculus]:

• When $x = 2$: $u = (2)^2 + 1 = 5$
• When $x = 3$: $u = (3)^2 + 1 = 10$

Step 4: Evaluating the Integral

Substituting the variables into the integral expression:

$$I = \int_{5}^{10} \frac{1}{2} \cdot \frac{1}{u} du$$

Extracting the constant factor and applying the standard integral formula $\int \frac{1}{u} du = \ln|u| + C$:

$$I = \frac{1}{2} [\ln|u|]_{5}^{10}$$

Applying the limits [Fundamental Theorem of Calculus, Part II]:

$$I = \frac{1}{2} (\ln 10 - \ln 5)$$

Step 5: Applying Logarithmic Identities

Using the property $\ln a - \ln b = \ln(\frac{a}{b})$:

$$I = \frac{1}{2} \ln\left(\frac{10}{5}\right)$$

$$I = \frac{1}{2} \ln(2)$$

Visual Representation of the Function

Final Solution: \frac{1}{2} \ln 2

Step 1: Theoretical Foundation

To evaluate the definite integral $\int_{a}^{b} f(x) dx$, we utilize the Fundamental Theorem of Calculus (Part II), which states:

$\int_{a}^{b} f(x) dx = F(b) - F(a)$

where $F(x)$ is the anti-derivative of the function $f(x)$ [Per the Fundamental Theorem of Calculus].

Step 2: Identifying the Anti-Derivative

The integrand is $f(x) = \frac{1}{x}$. The standard integral for the reciprocal function is given by:

$\int \frac{1}{x} dx = \ln|x| + C$

Thus, our anti-derivative function is $F(x) = \ln|x|$.

Step 3: Application of Limits

Given the definite integral with limits $a = 2$ and $b = 3$:

$\int_{2}^{3} \frac{1}{x} dx = \left[ \ln|x| \right]_{2}^{3}$

Applying the upper and lower limits [By the definition of definite integrals]:

$= \ln|3| - \ln|2|$

Step 4: Logarithmic Simplification

Using the quotient rule for logarithms, which states $\ln(m) - \ln(n) = \ln\left(\frac{m}{n}\right)$ for $m, n > 0$ [Per Logarithmic Identity Laws]:

$= \ln\left(\frac{3}{2}\right)$

Visual Representation of the Integrand Area

Final Calculation

The numerical value is the natural logarithm of $1.5$.

Final Solution: $\ln\left(\frac{3}{2}\right)$ or $\ln(1.5)$

Step 1: Decomposition of the Integrand

Given the integral $I = \int_{1}^{2} \frac{5x^2}{x^2+4x+3} dx$. Since the degree of the numerator is equal to the degree of the denominator, we perform long division [Polynomial Division Theorem].

Divide $5x^2$ by $x^2 + 4x + 3$:

Thus, the integrand is expressed as:

$\frac{5x^2}{x^2+4x+3} = 5 - \frac{20x + 15}{x^2 + 4x + 3}$

Step 2: Partial Fraction Decomposition

Factorize the denominator: $x^2 + 4x + 3 = (x+1)(x+3)$. We set up the partial fractions for the remainder term:

$\frac{20x + 15}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3}$

Multiply by $(x+1)(x+3)$: $20x + 15 = A(x+3) + B(x+1)$.

• Set $x = -1$: $20(-1) + 15 = A(-1+3) \implies -5 = 2A \implies A = -\frac{5}{2}$
• Set $x = -3$: $20(-3) + 15 = B(-3+1) \implies -45 = -2B \implies B = \frac{45}{2}$

Step 3: Setting up the Definite Integral

Substitute back into the integral expression:

$I = \int_{1}^{2} 5 \, dx - \int_{1}^{2} \left( \frac{-5/2}{x+1} + \frac{45/2}{x+3} \right) \, dx$

$I = \int_{1}^{2} 5 \, dx + \frac{5}{2} \int_{1}^{2} \frac{1}{x+1} \, dx - \frac{45}{2} \int_{1}^{2} \frac{1}{x+3} \, dx$

Step 4: Integration and Application of Limits

Apply the Fundamental Theorem of Calculus [$\int \frac{1}{x} dx = \ln|x|$]:

$I = [5x]_{1}^{2} + \frac{5}{2} [\ln|x+1|]_{1}^{2} - \frac{45}{2} [\ln|x+3|]_{1}^{2}$

$I = 5(2 - 1) + \frac{5}{2} (\ln 3 - \ln 2) - \frac{45}{2} (\ln 5 - \ln 4)$

Step 5: Simplification

$I = 5 + \frac{5}{2} \ln(\frac{3}{2}) - \frac{45}{2} \ln(\frac{5}{4})$

Using logarithm properties [$\ln a - \ln b = \ln(a/b)$ and $n \ln a = \ln a^n$]:

$I = 5 + \ln \left( \left(\frac{3}{2}\right)^{5/2} \right) - \ln \left( \left(\frac{5}{4}\right)^{45/2} \right)$

$I = 5 + \frac{5}{2} \ln 3 - \frac{5}{2} \ln 2 - \frac{45}{2} \ln 5 + \frac{45}{2} \ln 4$

Since $\ln 4 = 2 \ln 2$:

$I = 5 + \frac{5}{2} \ln 3 - \frac{5}{2} \ln 2 - \frac{45}{2} \ln 5 + 45 \ln 2$

$I = 5 + \frac{5}{2} \ln 3 + \frac{85}{2} \ln 2 - \frac{45}{2} \ln 5$

Final Solution: $5 + \frac{5}{2} \ln 3 + \frac{85}{2} \ln 2 - \frac{45}{2} \ln 5$ or $5 + \frac{5}{2} \ln\left( \frac{3 \cdot 2^{17}}{5^9} \right)$

Step 1: Application of the Fundamental Theorem of Calculus

To evaluate the definite integral of the polynomial function $f(x) = 4x^3 - 5x^2 + 6x + 9$ over the interval $[1, 2]$, we utilize the Second Fundamental Theorem of Calculus, which states that if $F(x)$ is the antiderivative of $f(x)$, then:

$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$ [Per the Fundamental Theorem of Calculus, Part II]

Step 2: Determining the Antiderivative

We apply the power rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, to each term of the integrand individually:

• $\int 4x^3 \, dx = 4 \left( \frac{x^4}{4} \right) = x^4$
• $\int -5x^2 \, dx = -5 \left( \frac{x^3}{3} \right) = -\frac{5}{3}x^3$
• $\int 6x \, dx = 6 \left( \frac{x^2}{2} \right) = 3x^2$
• $\int 9 \, dx = 9x$

Thus, the antiderivative is $F(x) = x^4 - \frac{5}{3}x^3 + 3x^2 + 9x$.

Step 3: Evaluating at Boundaries

We evaluate $F(x)$ at the upper bound ($x=2$) and lower bound ($x=1$):

For $x = 2$:

$F(2) = (2)^4 - \frac{5}{3}(2)^3 + 3(2)^2 + 9(2)$

$F(2) = 16 - \frac{5}{3}(8) + 3(4) + 18$

$F(2) = 16 - \frac{40}{3} + 12 + 18 = 46 - \frac{40}{3}$

$F(2) = \frac{138 - 40}{3} = \frac{98}{3}$

For $x = 1$:

$F(1) = (1)^4 - \frac{5}{3}(1)^3 + 3(1)^2 + 9(1)$

$F(1) = 1 - \frac{5}{3} + 3 + 9 = 13 - \frac{5}{3}$

$F(1) = \frac{39 - 5}{3} = \frac{34}{3}$

Step 4: Calculating the Definite Integral

Subtracting $F(1)$ from $F(2)$:

$\int_{1}^{2} f(x) \, dx = \frac{98}{3} - \frac{34}{3}$

$\int_{1}^{2} f(x) \, dx = \frac{64}{3}$

Visualization of the Definite Integral

Final Solution: \frac{64}{3}

Step 1: Setting up the Integral and Identifying the Method

We are tasked with evaluating the definite integral:

$I = \int_{0}^{1} x e^{x^2} dx$

Given the integrand is a composition of functions where the derivative of the exponent ($x^2$) is proportional to the multiplier ($x$), we employ the Method of Substitution [Per the substitution rule for definite integrals].

Step 2: Performing Algebraic Substitution

Let $u = x^2$.

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = 2x \implies du = 2x dx \implies x dx = \frac{1}{2} du$

Step 3: Transforming the Limits of Integration

We must adjust the limits of integration from the variable $x$ to the variable $u$ using the relation $u = x^2$:

• When $x = 0$, $u = (0)^2 = 0$.
• When $x = 1$, $u = (1)^2 = 1$.

[By the Fundamental Theorem of Calculus, the transformation of bounds is required when performing change of variables].

Step 4: Substitution and Evaluation

Substituting $u$ and $du$ into the original integral:

$I = \int_{0}^{1} e^u \left( \frac{1}{2} du \right)$

$I = \frac{1}{2} \int_{0}^{1} e^u du$

Evaluating the antiderivative [Per the integration rule $\int e^u du = e^u + C$]:

$I = \frac{1}{2} [e^u]_{0}^{1}$

$I = \frac{1}{2} (e^1 - e^0)$

Step 5: Final Simplification

Since $e^0 = 1$ [By the zero exponent rule in algebra]:

$I = \frac{1}{2} (e - 1)$

Visual Representation of the Area under the Curve

Final Solution: \frac{e-1}{2}

Step 1: Identification of the Integral Form

The given definite integral is:

$I = \int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}$

This integral corresponds to the standard derivative form of the inverse trigonometric function $\arcsin(x)$. [Per the Fundamental Theorem of Calculus and standard integration tables, $\int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right) + C$.]

Step 2: Determining the Antiderivative

In this specific case, $a = 1$. Therefore, the antiderivative of $f(x) = \frac{1}{\sqrt{1-x^2}}$ is:

$F(x) = \sin^{-1}(x)$

By the Second Fundamental Theorem of Calculus, if $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$, then:

$\int_{a}^{b} f(x) dx = F(b) - F(a)$

Step 3: Evaluating the Definite Integral

Substituting the limits $0$ and $1$ into the antiderivative $F(x) = \sin^{-1}(x)$:

$I = [\sin^{-1}(x)]_{0}^{1}$

$I = \sin^{-1}(1) - \sin^{-1}(0)$

Step 4: Trigonometric Evaluation

We evaluate the inverse sine values based on the unit circle definitions:

• $\sin^{-1}(1) = \frac{\pi}{2}$ [Since $\sin(\frac{\pi}{2}) = 1$ and $\frac{\pi}{2} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$]
• $\sin^{-1}(0) = 0$ [Since $\sin(0) = 0$ and $0 \in [-\frac{\pi}{2}, \frac{\pi}{2}]$]

$I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$

Visual Representation of the Function

Final Solution: \frac{\pi}{2}

Step 1: Identification of the Integral Form

The given definite integral is:

$I = \int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}$

We observe that the integrand is of the standard form $\frac{1}{a^2 + x^2}$, where $a = 1$. [Per the standard integral table for inverse trigonometric functions]

Step 2: Applying the Fundamental Theorem of Calculus

We recall the standard result for the antiderivative:

$\int \frac{dx}{1+x^2} = \tan^{-1}(x) + C$

Applying the Fundamental Theorem of Calculus, Part II, which states that if $F$ is the antiderivative of $f$, then $\int_{a}^{b} f(x) dx = F(b) - F(a)$:

$I = [\tan^{-1}(x)]_{1}^{\sqrt{3}}$

Step 3: Evaluation of the Definite Integral

Substitute the upper and lower limits into the evaluated antiderivative:

$I = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1)$

Step 4: Calculation of Trigonometric Values

Using the principal values of the inverse tangent function:

• $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$ [Since $\tan(\frac{\pi}{3}) = \sqrt{3}$]
• $\tan^{-1}(1) = \frac{\pi}{4}$ [Since $\tan(\frac{\pi}{4}) = 1$]

Therefore:

$I = \frac{\pi}{3} - \frac{\pi}{4}$

Step 5: Final Algebraic Simplification

To subtract the fractions, find the least common denominator (LCD), which is $12$:

$I = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{\pi}{12}$

Final Solution: \frac{\pi}{12}

Step 1: Establishing the Theoretical Framework

The given integral is $\int_{2}^{3} \frac{dx}{x^2 - 1}$. We recognize that the denominator is a difference of squares, $x^2 - 1 = (x - 1)(x + 1)$. This indicates the application of the method of Partial Fraction Decomposition [Per the algebraic identity for integration of rational functions].

Step 2: Partial Fraction Decomposition

We decompose the integrand into simpler fractions:

$\frac{1}{x^2 - 1} = \frac{1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$

Multiplying both sides by $(x-1)(x+1)$ yields:

$1 = A(x + 1) + B(x - 1)$

• Setting $x = 1$: $1 = A(2) \Rightarrow A = \frac{1}{2}$
• Setting $x = -1$: $1 = B(-2) \Rightarrow B = -\frac{1}{2}$

Thus, the integral becomes: $\int_{2}^{3} \left( \frac{1/2}{x-1} - \frac{1/2}{x+1} \right) dx$.

Step 3: Integration and Evaluation

Using the linearity property of integration [Per the Linearity Property of Definite Integrals]:

$\frac{1}{2} \int_{2}^{3} \frac{1}{x-1} dx - \frac{1}{2} \int_{2}^{3} \frac{1}{x+1} dx$

Evaluating the antiderivatives [Per $\int \frac{1}{u} du = \ln|u| + C$]:

$= \frac{1}{2} \left[ \ln|x-1| - \ln|x+1| \right]_{2}^{3}$

$= \frac{1}{2} \left[ \ln \left| \frac{x-1}{x+1} \right| \right]_{2}^{3}$

Step 4: Applying Fundamental Theorem of Calculus

Applying the limits of integration [Per the Fundamental Theorem of Calculus Part II]:

$= \frac{1}{2} \left( \ln \left| \frac{3-1}{3+1} \right| - \ln \left| \frac{2-1}{2+1} \right| \right)$

$= \frac{1}{2} \left( \ln \left| \frac{2}{4} \right| - \ln \left| \frac{1}{3} \right| \right)$

$= \frac{1}{2} \left( \ln \left( \frac{1}{2} \right) - \ln \left( \frac{1}{3} \right) \right)$

Step 5: Simplifying using Logarithmic Properties

Applying the quotient rule for logarithms [Per $\ln(a) - \ln(b) = \ln(a/b)$]:

$= \frac{1}{2} \ln \left( \frac{1/2}{1/3} \right) = \frac{1}{2} \ln \left( \frac{3}{2} \right)$

Visual Representation of the Integrand

Final Solution: $\frac{1}{2} \ln \left( \frac{3}{2} \right)$

Step 1: Theoretical Foundation

The integral to be evaluated is defined as $I = \int_{0}^{1} \frac{dx}{1+x^2}$. This integral follows the standard form of the derivative of the inverse trigonometric function, specifically the arctangent function [Per the Fundamental Theorem of Calculus and standard integration tables].

The antiderivative of $\frac{1}{1+x^2}$ with respect to $x$ is defined as:

$\int \frac{dx}{1+x^2} = \tan^{-1}(x) + C$

Step 2: Applying the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if $f$ is a continuous function on $[a, b]$ and $F$ is an antiderivative of $f$, then:

$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$

In this specific case, $a = 0$, $b = 1$, and $F(x) = \tan^{-1}(x)$.

Step 3: Execution of Limits

Substituting the upper and lower limits into the antiderivative:

$I = \left[ \tan^{-1}(x) \right]_{0}^{1}$

$I = \tan^{-1}(1) - \tan^{-1}(0)$

Step 4: Trigonometric Evaluation

Evaluating the inverse tangent values based on unit circle properties:

• $\tan^{-1}(1)$: The angle $\theta$ such that $\tan(\theta) = 1$ is $\frac{\pi}{4}$.
• $\tan^{-1}(0)$: The angle $\theta$ such that $\tan(\theta) = 0$ is $0$.

Substituting these values:

$I = \frac{\pi}{4} - 0 = \frac{\pi}{4}$

Geometric Representation

Analytical Summary

Final Solution: \frac{\pi}{4}

Step 1: Identifying the Standard Integral Form

The given integral is of the form $\int \frac{dx}{a^2 + f(x)^2}$. To solve this, we utilize the standard integration formula:

$\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$

Given the expression $\int_{0}^{\frac{2}{3}} \frac{dx}{4+9x^2}$, we observe that the denominator is not in the form $a^2 + x^2$. We must manipulate the denominator to isolate the $x^2$ term.

Step 2: Algebraic Manipulation of the Integrand

Factor out the constant coefficient of $x^2$ from the denominator:

$4 + 9x^2 = 9\left(\frac{4}{9} + x^2\right)$

Substituting this back into the integral:

$I = \int_{0}^{\frac{2}{3}} \frac{dx}{9\left(\left(\frac{2}{3}\right)^2 + x^2\right)} = \frac{1}{9} \int_{0}^{\frac{2}{3}} \frac{dx}{\left(\frac{2}{3}\right)^2 + x^2}$

Step 3: Application of the Integration Formula

Using the identity $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$, where $a = \frac{2}{3}$:

$I = \frac{1}{9} \left[ \frac{1}{2/3} \tan^{-1}\left(\frac{x}{2/3}\right) \right]_{0}^{\frac{2}{3}}$

[By the Fundamental Theorem of Calculus, $\int_{a}^{b} f(x)dx = F(b) - F(a)$]

Step 4: Evaluation of Definite Integral Limits

Simplify the coefficient: $\frac{1}{9} \times \frac{3}{2} = \frac{3}{18} = \frac{1}{6}$

$I = \frac{1}{6} \left[ \tan^{-1}\left(\frac{3x}{2}\right) \right]_{0}^{\frac{2}{3}}$

Substitute the upper and lower limits:

$I = \frac{1}{6} \left[ \tan^{-1}\left(\frac{3}{2} \cdot \frac{2}{3}\right) - \tan^{-1}\left(\frac{3}{2} \cdot 0\right) \right]$

$I = \frac{1}{6} \left[ \tan^{-1}(1) - \tan^{-1}(0) \right]$

Step 5: Final Trigonometric Calculation

Given the standard values of inverse trigonometric functions:

• $\tan^{-1}(1) = \frac{\pi}{4}$
• $\tan^{-1}(0) = 0$

$I = \frac{1}{6} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{24}$

Final Solution: \frac{\pi}{24}