Question

The value of the integral \(\int_{0}^{2} x^2 (2 - x)^5 \, dx\) is:

• A. $\frac{128}{21}$
• B. $\frac{64}{7}$
• C. $\frac{32}{21}$
• D. $\frac{16}{7}$

Hint:

For integrals of the form $\int_{0}^{a} x^m (a-x)^n dx$, you can use the shortcut: \[ \text{Result} = a^{m+n+1} \frac{m! n!}{(m+n+1)!} \] Here, $2^{2+5+1} \frac{2! 5!}{8!} = 256 \cdot \frac{2 \times 120}{40320} = \frac{256}{168} = \frac{32}{21}$.

Answer 1

The Correct Option is C

Concept: Based on the problem, we utilize the specific property of definite integrals: $$\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$$ This substitution simplifies the term $(2-x)^5$ into $x^5$, making the polynomial expansion much easier to integrate. Alternatively, one can use the Beta function identity for $\int_{0}^{a} x^m (a-x)^n dx$.

Step 1: Applying the integral property.
Let $I = \int_{0}^{2} x^2 (2 - x)^5 \, dx$. Applying the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$, we substitute $x$ with $(2 - x)$: \[ I = \int_{0}^{2} (2 - x)^2 (2 - (2 - x))^5 \, dx \] \[ I = \int_{0}^{2} (4 - 4x + x^2) x^5 \, dx \]

Step 2: Expanding and integrating.
Distribute $x^5$ into the parentheses: \[ I = \int_{0}^{2} (4x^5 - 4x^6 + x^7) \, dx \] Integrating term by term: \[ I = \left[ 4\frac{x^6}{6} - 4\frac{x^7}{7} + \frac{x^8}{8} \right]_{0}^{2} \] \[ I = \left[ \frac{2}{3}x^6 - \frac{4}{7}x^7 + \frac{1}{8}x^8 \right]_{0}^{2} \]

Step 3: Calculating the final value.
Substitute the upper limit $x=2$: \[ I = \frac{2}{3}(64) - \frac{4}{7}(128) + \frac{1}{8}(256) \] \[ I = \frac{128}{3} - \frac{512}{7} + 32 \] Using a common denominator of 21: \[ I = \frac{128(7) - 512(3) + 32(21)}{21} \] \[ I = \frac{896 - 1536 + 672}{21} = \frac{32}{21} \]