Find \int_1^2\sqrt{x}-3x \,dx.

Category: QuestionsFind \int_1^2\sqrt{x}-3x \,dx.
Editor">Editor Staff asked 11 months ago

Find \int_1^2\sqrt{x}-3x \,dx.
 
(a) \frac{8\sqrt{2}-31}{6}
 
(b) 8\sqrt{2}-31
 
(c) \frac{\sqrt{2}-31}{3}
 
(d) \frac{8\sqrt{2}+31}{4}
 
This question was posed to me in an interview for job.
 
My query is from Fundamental Theorem of Calculus-2 topic in section 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 answer is (a) \frac{8\sqrt{2}-31}{6}
 
The best I can explain: Let I=\int_1^2 \sqrt{x}-3x \,dx
 
F(x)=\int \sqrt{x}-3x \,dx
 
=\frac{x^{1/2+1}}{1/2+1}-\frac{3x^2}{2}=\frac{2x^{\frac{3}{2}}}{3}-\frac{3x^2}{2}
 
By using the second fundamental theorem of calculus, we get
 
I=F(2)-F(1)=\left(\frac{2×2^{3/2}}{3}-\frac{3×2^2}{2}\right)-\left(\frac{2×1^{3/2}}{3}-\frac{3×1^2}{2}\right)
 
I=\frac{4\sqrt{2}}{3}-6-\frac{2}{3}+\frac{3}{2}=\frac{8\sqrt{2}-36-4+9}{6}=\frac{8\sqrt{2}-31}{6}