Question:
The value of the integral $I=\int\limits^{6}_{{3}}$$\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx$ is :
The value of the integral $I=\int\limits^{6}_{{3}}$$\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx$ is :
Updated On: Jul 27, 2022
- $\frac{3}{2}$
- $2$
- $1$
- $\frac{1}{2}$
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The Correct Option is A
Solution and Explanation
Let $I=\int\limits^{6}_{{3}}$$\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx ...\left(i\right)$
$=\int\limits^{6}_{{3}}$$\frac{\sqrt{9-x}}{\sqrt{9-9+x}+\sqrt{9-x}}dx$
$\Rightarrow$ $I=\int\limits^{6}_{{3}}$ $\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx ...\left(ii\right)$
On adding Eqs. (i) and (ii), we get
$2I=\int\limits^{6}_{{3}}$$\frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx$
$=\int\limits^{6}_{{3}}$$1\,dx=\left[x\right]_{3}^{6}$
$=6-3=3$
$\Rightarrow I=\frac{3}{2}$
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.