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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1 Answers
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}