Question

The value of $\int_0^2 x(x^2 + 1)^3 \, dx$ is given by

• A. 77
• B. 76
• C. 78
• D. 79

Hint:

Always change the limits of integration immediately during substitution. This prevents the extra step of back-substituting the $x$-terms at the end.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To evaluate a definite integral, we can apply substitution and adjust the limits of integration accordingly.
Key Formula or Approach:
If we let $u = g(x)$, then:
\[ \int_a^b f(g(x))g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du \]

Step 2: Detailed Explanation:
Let the integral be:
\[ I = \int_0^2 x(x^2 + 1)^3 \, dx \]
Use the substitution:
\[ u = x^2 + 1 \implies du = 2x \, dx \implies x \, dx = \frac{du}{2} \]
Now, calculate the new limits of integration for $u$:
- Lower limit: when $x = 0 \implies u = 0^2 + 1 = 1$
- Upper limit: when $x = 2 \implies u = 2^2 + 1 = 5$
Substitute these into the integral:
\[ I = \int_1^5 u^3 \cdot \frac{du}{2} \]
\[ I = \frac{1}{2} \int_1^5 u^3 \, du \]
Evaluate the integral:
\[ I = \frac{1}{2} \left[ \frac{u^4}{4} \right]_1^5 \]
\[ I = \frac{1}{8} \left[ 5^4 - 1^4 \right] \]
\[ I = \frac{1}{8} [625 - 1] = \frac{624}{8} = 78 \]
Therefore, the value of the definite integral is 78.

Step 3: Final Answer
The correct option is (C).