Question

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

Answer 1

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