Question

Evaluate the integral: $$ \int_0^2 x^2 (2 - x)^5 \, dx $$

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

Hint:

Use substitution to simplify integral limits and integrand, then integrate term-wise.

Answer 1

The Correct Option is C

Let \( I = \int_0^2 x^2 (2 - x)^5 \, dx \). Substitute \( t = 2 - x \), then \( dx = -dt \) and when \( x=0, t=2 \), when \( x=2, t=0 \). Rewrite integral: \[ I = \int_{t=2}^{0} (2 - t)^2 t^5 (-dt) = \int_0^2 (2 - t)^2 t^5 dt \] Expand \( (2 - t)^2 = 4 - 4t + t^2 \): \[ I = \int_0^2 (4 - 4t + t^2) t^5 dt = \int_0^2 (4t^5 - 4t^6 + t^7) dt \] Integrate term-wise: \[ I = \left[ \frac{4 t^6}{6} - \frac{4 t^7}{7} + \frac{t^8}{8} \right]_0^2 = \left[ \frac{2}{3} t^6 - \frac{4}{7} t^7 + \frac{1}{8} t^8 \right]_0^2 \] Evaluate at \( t=2 \): \[ \frac{2}{3} \times 64 - \frac{4}{7} \times 128 + \frac{1}{8} \times 256 = \frac{128}{3} - \frac{512}{7} + 32 \] Convert all to common denominator 21: \[ \frac{128 \times 7}{21} - \frac{512 \times 3}{21} + \frac{32 \times 21}{21} = \frac{896}{21} - \frac{1536}{21} + \frac{672}{21} = \frac{896 - 1536 + 672}{21} = \frac{32}{21} \]