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

Category: QuestionsEvaluate the integral \int_1^6 \frac{\sqrt{x}+3}{\sqrt{x}} \,dx.
Editor">Editor Staff asked 11 months ago

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}

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This question is from Evaluation of Definite Integrals by Substitution in chapter Integrals of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

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}