# Find $$\int_1^2 log⁡x.x^2 dx$$

Category: QuestionsFind $$\int_1^2 log⁡x.x^2 dx$$
Editor">Editor Staff asked 11 months ago

Find $$\int_1^2 log⁡x.x^2 dx$$

(a) log⁡2-$$\frac{7}{3}$$

(b) $$\frac{8}{3}$$ log⁡2-5

(c) $$\frac{8}{3}$$ log⁡2-log⁡3

(d) $$\frac{8}{3}$$ log⁡2

I got this question in an interview for job.

Question is taken 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
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience

Editor">Editor Staff answered 11 months ago

Correct answer is (b) \frac{8}{3} log⁡2-5

To explain I would say: I=\int_0^1 log⁡x.x^2 dx

F(x)=\int log⁡x.x^2 dx

By using the formula \int u.v dx=u\int v dx-\int u'(\int v dx), we get

\int log⁡x.x^2 \,dx=log⁡x \int x^2 dx-\int (log⁡x)’ \int \,x^2 dx

=\frac{x^3 log⁡x}{3}-\int \frac{1}{x}.x^3/3 dx

∴F(x)=\frac{x^3 log⁡x}{3}-\frac{x^3}{9}=\frac{x^3}{3} (log⁡x-\frac{1}{3})

Hence, by using the fundamental theorem of calculus, we get

I=F(2)-F(1)

I=\frac{2^3}{3} \,(log⁡2-\frac{2}{3})-\frac{1^3}{3} \,(log⁡1-\frac{1}{3})

I=\frac{2^3}{3} \,log⁡2-\frac{16}{3}+\frac{1}{3}

I=\frac{8}{3} log⁡2-5