Question

If \(\int _0^2x(2-x)^b\,dx = \frac{32}{7}\), where \(b\in N\) then \(b =\)

• A. \(5\)
• B. \(6\)
• C. \(7\)
• D. \(8\)

Hint:

Use the property of replacing x by 2-x, then integrate and test values of b.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The integrand has the awkward factor \((2-x)^b\). Using the property \(\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx\) moves the power to \(x\).

Step 2: Key Formula or Approach:
\[ \int_0^2 x(2-x)^b\,dx = \int_0^2 (2-x)\,x^b\,dx \]

Step 3: Detailed Explanation:
Expand and integrate:
\[ \int_0^2 \left(2x^b - x^{b+1}\right)dx = \left[\frac{2x^{b+1}}{b+1} - \frac{x^{b+2}}{b+2}\right]_0^2 = \frac{2^{b+2}}{b+1} - \frac{2^{b+2}}{b+2} \]
\[ = 2^{b+2}\cdot\frac{1}{(b+1)(b+2)} \]
Set this equal to \(\dfrac{32}{7}\). Try \(b = 6\):
\[ \frac{2^8}{7\cdot8} = \frac{256}{56} = \frac{32}{7} \]
It matches. For the other options: \(b = 5\) gives \(\dfrac{128}{42} = \dfrac{64}{21}\), \(b = 7\) gives \(\dfrac{512}{72} = \dfrac{64}{9}\), and \(b = 8\) gives \(\dfrac{1024}{90}\), none equal to \(\dfrac{32}{7}\).

Final Answer:
\(b = 6\), option (B). \[ \boxed{6 \text{ (B)}} \]