1.

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»

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


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