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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
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}.