If ∫_-1^b x^3 dx=15/4 , then b can take the value

The correct answer is Option (3) → 2

    \( \displaystyle \int_{-1}^{b}x^3\,dx=\left[\frac{x^4}{4}\right]_{-1}^{b} \)

    \( \displaystyle =\frac{b^4}{4}-\frac{1}{4} \)

    \( \displaystyle \frac{b^4-1}{4}=\frac{15}{4} \)

    \( \displaystyle b^4=16 \)

So b = 2 or b = −2. Of the options, b = 2; this is why the question says b “can take” the value.