Step 1: Defining the Function and Region of Integration

The function is defined as $y = f(x) = x|x|$. Based on the definition of the absolute value function, we split the function into two intervals:

• If $x \geq 0$, then $|x| = x$, so $y = x(x) = x^2$.
• If $x < 0$, then $|x| = -x$, so $y = x(-x) = -x^2$.

The total area $A$ bounded by the curve, the $x$-axis, and the ordinates $x = -1$ and $x = 1$ is given by the definite integral of the absolute value of the function over the interval $[-1, 1]$:

$A = \int_{-1}^{1} |f(x)| \, dx = \int_{-1}^{1} |x|x|| \, dx$

Since $|x|x|| = |x^2| = x^2$ for all real $x$, the area simplifies to:

$A = \int_{-1}^{1} x^2 \, dx$

Step 2: Evaluating the Integral using Symmetry

The function $g(x) = x^2$ is an even function, meaning $g(-x) = (-x)^2 = x^2 = g(x)$. [Per the property of definite integrals for even functions: $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$].

Applying this property:

$A = 2 \int_{0}^{1} x^2 \, dx$

Step 3: Calculating the Definite Integral

Applying the Fundamental Theorem of Calculus: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$.

$A = 2 \left[ \frac{x^3}{3} \right]_{0}^{1}$

$A = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right)$

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

Visual Representation

Summary of Calculation

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

Final Solution: The area bounded by the curve is $\frac{2}{3}$ square units.

Step 1: Identifying the Geometric Properties

The given equation is $\frac{x^2}{16} + \frac{y^2}{9} = 1$. This is the standard form of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 16$ and $b^2 = 9$.

• Semi-major axis ($a$): $\sqrt{16} = 4$ units.
• Semi-minor axis ($b$): $\sqrt{9} = 3$ units.
• Center: $(0,0)$.

Step 2: Graphical Representation

Step 3: Setting up the Integral

Due to the symmetry of the ellipse about both the $x$-axis and $y$-axis, the total area $A$ is four times the area of the region in the first quadrant. [By Symmetry Principle]

From $\frac{x^2}{16} + \frac{y^2}{9} = 1$, we isolate $y$ for the first quadrant ($y \ge 0$):

$\frac{y^2}{9} = 1 - \frac{x^2}{16} \implies y^2 = \frac{9}{16}(16 - x^2) \implies y = \frac{3}{4}\sqrt{16 - x^2}$

The area in the first quadrant is $\int_{0}^{4} y \, dx$. Thus, the total area $A$ is:

$A = 4 \int_{0}^{4} \frac{3}{4} \sqrt{4^2 - x^2} \, dx = 3 \int_{0}^{4} \sqrt{4^2 - x^2} \, dx$

Step 4: Evaluating the Definite Integral

Using the standard integral formula: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})$

$A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2} \sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4}$

$A = 3 \left[ \left( \frac{4}{2}\sqrt{16 - 16} + 8 \sin^{-1}(1) \right) - \left( 0 + 8 \sin^{-1}(0) \right) \right]$

$A = 3 \left[ 0 + 8\left(\frac{\pi}{2}\right) - 0 \right]$

$A = 3 \times 4\pi = 12\pi$

Summary of Calculation

Final Solution: The area of the region bounded by the ellipse is $12\pi$ square units.

Step 1: Theoretical Foundation & Interval Analysis

The area $A$ bounded by a curve $y = f(x)$, the $x$-axis, and the ordinates $x = a$ and $x = b$ is given by the definite integral: $A = \int_{a}^{b} |f(x)| \, dx$ [Per the Fundamental Theorem of Calculus].

Given the function $f(x) = x^3$ and the interval $[-2, 1]$, we observe that the function changes sign at $x = 0$.

• For $x \in [-2, 0]$, $f(x) \le 0$ (the curve lies below the $x$-axis).
• For $x \in [0, 1]$, $f(x) \ge 0$ (the curve lies above the $x$-axis).

Step 2: Geometrical Representation

Step 3: Integration and Evaluation

Applying the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ [Fundamental Theorem of Calculus]:

Part A: Integral from -2 to 0 $I_1 = \int_{-2}^{0} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{-2}^{0}$ $I_1 = \left( \frac{0^4}{4} \right) - \left( \frac{(-2)^4}{4} \right) = 0 - \frac{16}{4} = -4$ Area $A_1 = |I_1| = |-4| = 4 \text{ square units}$.

Part B: Integral from 0 to 1 $I_2 = \int_{0}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{1}$ $I_2 = \left( \frac{1^4}{4} \right) - \left( \frac{0^4}{4} \right) = \frac{1}{4} - 0 = 0.25$ Area $A_2 = |I_2| = |0.25| = 0.25 \text{ square units}$.

Step 4: Total Summation

The total area bounded is: $A = A_1 + A_2 = 4 + 0.25 = 4.25$ Expressing as a fraction: $A = 4 + \frac{1}{4} = \frac{17}{4} \text{ square units}$.

Final Solution: The area bounded by the curve is \frac{17}{4} \text{ square units (or 4.25 square units).}

Step 1: Analyzing the Equation of the Ellipse

The given equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$. This is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where:

• $a^2 = 4 \implies a = 2$ (Semi-major axis along the $x$-axis is $2$)
• $b^2 = 9 \implies b = 3$ (Semi-major axis along the $y$-axis is $3$)

[Per standard conic section definitions, since $b > a$, the ellipse is elongated along the $y$-axis.]

Step 2: Expressing $y$ in terms of $x$

To find the area using integration, we isolate $y$ from the equation:

$\frac{y^2}{9} = 1 - \frac{x^2}{4}$

$y^2 = 9 \left(1 - \frac{x^2}{4}\right) = \frac{9}{4}(4 - x^2)$

$y = \pm \frac{3}{2}\sqrt{4 - x^2}$

Since the ellipse is symmetric about both axes, the total area $A$ is 4 times the area of the region in the first quadrant where $x$ ranges from $0$ to $2$.

Step 3: Setting up the Definite Integral

$A = 4 \int_{0}^{2} y \, dx = 4 \int_{0}^{2} \frac{3}{2} \sqrt{2^2 - x^2} \, dx$

$A = 6 \int_{0}^{2} \sqrt{2^2 - x^2} \, dx$

Step 4: Applying the Integration Formula

Using the standard integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$:

$A = 6 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

$A = 6 \left[ \left( \frac{2}{2}\sqrt{4 - 4} + 2\sin^{-1}(1) \right) - \left( 0 + 2\sin^{-1}(0) \right) \right]$

$A = 6 \left[ 0 + 2(\frac{\pi}{2}) - 0 \right]$

$A = 6 \pi$

Visual Representation of the Ellipse

Summary of Calculation

Final Solution: The area of the region bounded by the ellipse is $6\pi$ square units.

Step 1: Analyzing the Geometric Constraints

The given equation is $x^2 + y^2 = 4$, which represents a circle centered at the origin $(0, 0)$ with radius $r = \sqrt{4} = 2$ units [By the standard form of circle equation $(x-h)^2 + (y-k)^2 = r^2$].

The boundary conditions are:

• The circle: $y^2 = 4 - x^2 \implies y = \sqrt{4 - x^2}$ (taking the positive root as we are constrained to the first quadrant).
• The vertical lines: $x = 0$ (the $y$-axis) and $x = 2$ (the rightmost edge of the circle).

Step 2: Defining the Integral Setup

The area $A$ in the first quadrant bounded by the curve $f(x) = \sqrt{4 - x^2}$ from $x = 0$ to $x = 2$ is given by the definite integral:

$A = \int_{0}^{2} \sqrt{4 - x^2} \, dx$

Step 3: Evaluating the Integral

Using the standard integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$:

Substitute $a = 2$:

$A = \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2} \sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

$A = \left[ \frac{x}{2}\sqrt{4 - x^2} + 2 \sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

Applying the Fundamental Theorem of Calculus:

$A = \left( \frac{2}{2}\sqrt{4 - 2^2} + 2 \sin^{-1}\left(\frac{2}{2}\right) \right) - \left( \frac{0}{2}\sqrt{4 - 0^2} + 2 \sin^{-1}\left(\frac{0}{2}\right) \right)$

$A = (1 \cdot 0 + 2 \cdot \sin^{-1}(1)) - (0 + 2 \cdot 0)$

$A = 2 \cdot \left(\frac{\pi}{2}\right) = \pi$

Step 4: Geometric Visualization

Final Solution: The area is $\pi$ square units.

Given Variables & Initial Setup

The region is bounded by:

• The parabola: $y^2 = 4x$
• The $y$-axis: $x = 0$
• The horizontal line: $y = 3$

To find the area with respect to the $y$-axis, we express $x$ in terms of $y$ from the given equation of the parabola:

$x = \frac{y^2}{4}$ [By algebraic rearrangement]

Step 1: Establishing the Limits of Integration

The region starts from the origin where $y = 0$ and extends up to the line $y = 3$. Therefore, the definite integral limits are $[0, 3]$.

Step 2: Defining the Integral for Area

The area $A$ under the curve bounded by the $y$-axis is given by the integral of $x$ with respect to $y$:

$A = \int_{c}^{d} x \, dy = \int_{0}^{3} \frac{y^2}{4} \, dy$

Step 3: Calculating the Integral

Applying the Power Rule for integration, $\int y^n \, dy = \frac{y^{n+1}}{n+1}$:

$A = \frac{1}{4} \int_{0}^{3} y^2 \, dy$

$A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3}$ [By the Fundamental Theorem of Calculus]

$A = \frac{1}{4} \left( \frac{3^3}{3} - \frac{0^3}{3} \right)$

$A = \frac{1}{4} \left( \frac{27}{3} \right)$

$A = \frac{1}{4} (9)$

$A = \frac{9}{4} \text{ sq. units}$

Visual Representation

Final Solution: The area of the region is $\frac{9}{4}$ or $2.25 \text{ square units}$.

Step 1: Identifying the Region and Theoretical Setup

The problem asks for the area of the region bounded by the parabola $y = x^2$, the vertical lines $x = 1$ and $x = 2$, and the $x$-axis. Since the function $y = x^2$ is continuous and non-negative for all $x \in [1, 2]$, the area $A$ is given by the definite integral of the function $f(x)$ with respect to $x$ over the interval $[a, b]$.

Formula: $A = \int_{a}^{b} |f(x)| \, dx = \int_{1}^{2} x^2 \, dx$ [Per the Fundamental Theorem of Calculus].

Step 2: Performing the Integration

Apply the Power Rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n = 2$.

$$A = \int_{1}^{2} x^2 \, dx$$

$$A = \left[ \frac{x^{2+1}}{2+1} \right]_{1}^{2}$$

$$A = \left[ \frac{x^3}{3} \right]_{1}^{2}$$

Step 3: Applying Limits of Integration

Per the Second Fundamental Theorem of Calculus, substitute the upper limit and subtract the evaluation at the lower limit:

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

$$A = \frac{8}{3} - \frac{1}{3}$$

$$A = \frac{7}{3}$$

Summary of Calculation

Final Solution: The area under the curve is $\frac{7}{3}$ square units.

Given Variables & Initial Setup

We are tasked with determining the area bounded by the curve $y = f(x) = x^4$, the vertical lines $x = 1$ and $x = 5$, and the $x$-axis. Since the function $y = x^4$ is continuous and non-negative for all $x \in [1, 5]$, the area $A$ is given by the definite integral of the function with respect to $x$ over the specified interval [Per the Fundamental Theorem of Calculus].

Step 1: Formulating the Definite Integral

The area $A$ is defined as:

$A = \int_{a}^{b} y \, dx = \int_{1}^{5} x^4 \, dx$

Step 2: Performing the Integration

To evaluate the integral, we apply the Power Rule of Integration, which states that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$ [Power Rule of Calculus].

$A = \left[ \frac{x^{4+1}}{4+1} \right]_{1}^{5}$

$A = \left[ \frac{x^5}{5} \right]_{1}^{5}$

Step 3: Applying the Limits of Integration

Per the Second Fundamental Theorem of Calculus, we evaluate the antiderivative at the upper limit and subtract the evaluation at the lower limit:

$A = \left( \frac{5^5}{5} \right) - \left( \frac{1^5}{5} \right)$

$A = \frac{3125}{5} - \frac{1}{5}$

$A = \frac{3124}{5}$

Step 4: Final Numerical Calculation

$A = 624.8 \, \text{square units}$

Visual Representation

Final Solution: The area bounded by the curve is $\frac{3124}{5} = 624.8$ square units.

Step 1: Analyzing the Absolute Value Function

The given function is $y = |x + 3|$. By the definition of the absolute value function, $f(x) = |x + 3|$ is defined as:

$|x + 3| = \begin{cases} x + 3, & \text{if } x + 3 \geq 0 \implies x \geq -3 \\ -(x + 3), & \text{if } x + 3 < 0 \implies x < -3 \end{cases}$

[Per the definition of the Modulus Function]

Step 2: Visual Representation of the Function

Step 3: Setting up the Definite Integral

We are required to evaluate the integral $I = \int_{-6}^{0} |x + 3| \, dx$. Due to the change in the behavior of the function at the critical point $x = -3$, we apply the additive property of definite integrals [Per the Property of Definite Integrals $\int_a^c f(x)dx = \int_a^b f(x)dx + \int_b^c f(x)dx$]:

$I = \int_{-6}^{-3} -(x + 3) \, dx + \int_{-3}^{0} (x + 3) \, dx$

Step 4: Evaluating the Integral

Solving the first part:

$\int_{-6}^{-3} (-x - 3) \, dx = \left[ -\frac{x^2}{2} - 3x \right]_{-6}^{-3}$

$= \left( -\frac{(-3)^2}{2} - 3(-3) \right) - \left( -\frac{(-6)^2}{2} - 3(-6) \right)$

$= \left( -\frac{9}{2} + 9 \right) - \left( -\frac{36}{2} + 18 \right)$

$= (4.5) - (-18 + 18) = 4.5$

Solving the second part:

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

$= \left( \frac{0^2}{2} + 3(0) \right) - \left( \frac{(-3)^2}{2} + 3(-3) \right)$

$= 0 - \left( \frac{9}{2} - 9 \right) = 0 - (-4.5) = 4.5$

Step 5: Final Summation

$I = 4.5 + 4.5 = 9$

Alternatively, the integral represents the area of two congruent right-angled triangles with base $= 3$ and height $= 3$.

$\text{Area} = 2 \times \left( \frac{1}{2} \times \text{base} \times \text{height} \right) = 2 \times \left( \frac{1}{2} \times 3 \times 3 \right) = 9$

Final Solution: \int_{-6}^{0} |x + 3| \, dx = 9 \text{ square units}

Given Variables & Initial Setup

The area $A$ bounded by the curve $f(x) = \sin x$ between the limits $x = a$ and $x = b$ is defined by the definite integral of the absolute value of the function:

$A = \int_{0}^{2\pi} | \sin x | \, dx$

Step 1: Identifying Interval Sign Changes

The function $y = \sin x$ changes sign at the roots of the equation $\sin x = 0$ within the interval $[0, 2\pi]$. The roots are $x = 0, \pi, 2\pi$.

• For $x \in [0, \pi]$, $\sin x \ge 0$. [Per the unit circle definition in the first and second quadrants].
• For $x \in [\pi, 2\pi]$, $\sin x \le 0$. [Per the unit circle definition in the third and fourth quadrants].

Step 2: Splitting the Integral

By the property of additivity of definite integrals [Definite Integral Property: $\int_a^c f(x)dx = \int_a^b f(x)dx + \int_b^c f(x)dx$], we split the integral at $x = \pi$ to handle the absolute value:

$A = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} | \sin x | \, dx$

$A = \int_{0}^{\pi} \sin x \, dx - \int_{\pi}^{2\pi} \sin x \, dx$

Step 3: Performing the Integration

The antiderivative of $\sin x$ is $-\cos x$. [Per the Fundamental Theorem of Calculus].

Part 1: $\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$

Part 2: $-\int_{\pi}^{2\pi} \sin x \, dx = -[-\cos x]_{\pi}^{2\pi} = [\cos x]_{\pi}^{2\pi} = (\cos 2\pi - \cos \pi) = (1 - (-1)) = 2$

Step 4: Summation of Absolute Areas

$A = 2 + 2 = 4$

Visual Representation of the Bounded Area

Final Solution: The area bounded by the curve is 4 square units.