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

Category: QuestionsFind the value \int_{-1}^23x+x^2-2 \,dx.
Editor">Editor Staff asked 11 months ago

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
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Editor">Editor Staff answered 11 months ago

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}.