Question:
$\int \limits^{2017}_{2016} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{4033 - x}} dx $ is equal to
$\int \limits^{2017}_{2016} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{4033 - x}} dx $ is equal to
Updated On: Apr 8, 2024
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The Correct Option is D
Solution and Explanation
Let $I=\int\limits_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4033-x}} d x\,\,\,...(i)$
$\therefore I=\int\limits_{2016}^{2017} \frac{\sqrt{4033-x}}{\sqrt{4033-X}+\sqrt{4033-(4033-x)}} d x$
$\left[\because \int\limits_{a}^{b} f(x) d x=\int_{b}^{a} f(a+b-x) d x\right]$
$\Rightarrow I=\int\limits_{2016}^{2017} \frac{\sqrt{4033-X}}{\sqrt{4033-X}+\sqrt{x}} d x \ldots$ (ii)
On adding Eqs. (i) and (ii), we get
$2 I=\int\limits_{2016}^{2017} d x$
$\Rightarrow 2\, I=[x]_{2016}^{2017}$
$\Rightarrow 2 \,I=1$
$\Rightarrow I=\frac{1}{2}$
$\therefore I=\int\limits_{2016}^{2017} \frac{\sqrt{4033-x}}{\sqrt{4033-X}+\sqrt{4033-(4033-x)}} d x$
$\left[\because \int\limits_{a}^{b} f(x) d x=\int_{b}^{a} f(a+b-x) d x\right]$
$\Rightarrow I=\int\limits_{2016}^{2017} \frac{\sqrt{4033-X}}{\sqrt{4033-X}+\sqrt{x}} d x \ldots$ (ii)
On adding Eqs. (i) and (ii), we get
$2 I=\int\limits_{2016}^{2017} d x$
$\Rightarrow 2\, I=[x]_{2016}^{2017}$
$\Rightarrow 2 \,I=1$
$\Rightarrow I=\frac{1}{2}$
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
