Question

Evaluate the integral \(\int_1^6 \frac{\sqrt{x}+3}{\sqrt{x}} \,dx\).

(a) 9

(b) \(\frac{9}{2}\)

(c) –\(\frac{9}{2}\)

(d) \(\frac{4}{5}\)

I have been asked this question in an interview for internship.

This question is from Evaluation of Definite Integrals by Substitution in chapter Integrals of Mathematics – Class 12

Answer 1

The correct choice is (b) \(\frac{9}{2}\)

To explain: I=\(\int_1^4 \frac{\sqrt{x}+3}{\sqrt{x}} \,dx\)

Let \(\sqrt{x}+3=t\)

Differentiating w.r.t x, we get

\(\frac{1}{2\sqrt{x}} \,dx=dt\)

\(\frac{1}{\sqrt{x}} \,dx=2 \,dt\)

The new limits

When x=1,t=4

When x=4,t=5

∴\(\int_1^4 \frac{\sqrt{x}+3}{\sqrt{x}} dx=\int_4^5 \,t \,dt\)

=\([\frac{t^2}{2}]_4^5=\frac{5^2-4^2}{2}=\frac{9}{2}\)