Question:
The area bounded by the curve $y = \sin x$, $x$-axis and the ordinates $x = 0$ and $x = \pi /2$ is
The area bounded by the curve $y = \sin x$, $x$-axis and the ordinates $x = 0$ and $x = \pi /2$ is
Updated On: May 19, 2022
- $\pi$
- $\pi/2$
- 1
- 2
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The Correct Option is C
Solution and Explanation
Area $ = \int\limits^{\pi/2}_{0} y\, dx $
$= \int\limits^{\pi/2}_{0} \sin x \,dx$
$ = \left[-\cos x\right]^{\pi/2}_{0} = 1 $
$= \int\limits^{\pi/2}_{0} \sin x \,dx$
$ = \left[-\cos x\right]^{\pi/2}_{0} = 1 $
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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.
