\( I=\int_{2}^{8}\displaystyle\frac{\sqrt[3]{x+1}}{\sqrt[3]{x+1}+\sqrt[3]{11-x}}dx \) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
By the property \( \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx \) with a + b = 10, replacing x by 10 − x turns x + 1 into 11 − x, and 11 − x into x + 1:
\( \displaystyle I=\int_{2}^{8}\displaystyle\frac{\sqrt[3]{11-x}}{\sqrt[3]{11-x}+\sqrt[3]{x+1}}\,dx \)
Adding the two forms of I:
\( \displaystyle 2I=\int_{2}^{8}\displaystyle\frac{\sqrt[3]{x+1}+\sqrt[3]{11-x}}{\sqrt[3]{x+1}+\sqrt[3]{11-x}}\,dx \)
\( \displaystyle =\int_{2}^{8}\displaystyle 1\,dx \)
\( \displaystyle =6 \)
\( \displaystyle I=3 \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{2} \) — Incorrect. \( \frac{1}{2} \) is the value of the integrand at the middle point x = 5, not the integral.
Option (2) 6 — Incorrect. 6 is 2I (the length of the interval); it must be halved.
Option (3) 10 — Incorrect. 10 is a + b, not the value of the integral.