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

1 Answers
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=\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