Question

Find the value \(\int_{-1}^23x+x^2-2 \,dx\).

(a) –\(\frac{4}{3}\)

(b) \(\frac{3}{2}\)

(c) \(\frac{5}{6}\)

(d) –\(\frac{5}{6}\)

The question was posed to me during an online exam.

This intriguing question comes from Fundamental Theorem of Calculus-2 topic in portion Integrals of Mathematics – Class 12

Answer 1

Right choice is (b) \(\frac{3}{2}\)

Easiest explanation: Let \(I=\int_{-1}^23x+x^2-2 \,dx\)

F(x)=\(\int 3x+x^2-2 \,dx\)

=\(\frac{3x^2}{2}+\frac{x^3}{3}-2x\)

Applying the limits, we get

I=F(2)-F(-1)

I=\(\left(\frac{(3×2^3)}{2}+\frac{2^3}{3}-2(2)\right)-\left(\frac{3 (-1)^2}{2}+\frac{(-1)^3}{3}-2(-1)\right)\)

I=\(6+\frac{8}{3}-4-\frac{3}{2}+\frac{1}{3}-2=\frac{3}{2}\).