The value of \(\int_{-2}^{2}\displaystyle\frac{x^{2}}{1+3^{x}}\,dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{8}{3}\)
Let
\( \displaystyle I = \int_{-2}^{2}\frac{x^{2}}{1+3^{x}}\,dx \qquad (1) \)
By the property \(\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx\) with \(a+b = 0\), replace x by \(-x\):
\( \displaystyle I = \int_{-2}^{2}\frac{x^{2}}{1+3^{-x}}\,dx \)
Multiplying the numerator and denominator by \(3^{x}\),
\( \displaystyle I = \int_{-2}^{2}\frac{x^{2}\,3^{x}}{3^{x}+1}\,dx \qquad (2) \)
Adding (1) and (2), the fractions combine to \(\frac{1+3^{x}}{1+3^{x}} = 1\):
\( \displaystyle 2I = \int_{-2}^{2}x^{2}\,dx = \left[\frac{x^{3}}{3}\right]_{-2}^{2} = \frac{16}{3} \)
\( \displaystyle I = \frac{8}{3} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{16}{3}\) — Incorrect. This is 2I; it forgets to divide by 2 at the end.
Option (3) 0 — Incorrect. The integrand is not odd, and it is never negative, so the integral is positive.
Option (4) \(\displaystyle\frac{4}{3}\) — Incorrect. It integrates \(x^{2}\) over \([0, 2]\) only and halves that; the sum 2I is \(\int_{-2}^{2}x^{2}\,dx = \frac{16}{3}\) over the whole interval.