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Q3:
Evaluate the definite integrals in Exercises 1 to 20. 3. $\int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) dx$

Solution :

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

1 2 Area = 64/3

Final Solution: \frac{64}{3}


More Questions from Class 12 Mathematics Integrals EXERCISE 7.8


CBSE Solutions for Class 12 Mathematics Integrals


Chapters in CBSE - Class 12 Mathematics


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