Question:
Evaluate:$\int\limits_{0}^{2\pi} sin\left(\frac{\pi}{4}+\frac{x}{2}\right)dx $
Evaluate:$\int\limits_{0}^{2\pi} sin\left(\frac{\pi}{4}+\frac{x}{2}\right)dx $
Updated On: Jul 6, 2022
- $-2\sqrt{2}$
- $-2$
- $\sqrt{2}$
- $2\sqrt{2}$
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The Correct Option is D
Solution and Explanation
Let $I = \int\limits_{0}^{\frac{\pi}{2}} sin \left(\frac{\pi}{4} + \frac{x}{2}\right) dx$
$I= \left[-2\,cos\, \left(\frac{\pi}{4} +\,\frac{x}{2}\right)\right]_{0}^{2\pi} = -2 \left(cos\, \left(\frac{\pi}{4} + \pi\right) -cos\, \frac{\pi}{4}\right)$
$ = 2\,cos \,\frac{\pi}{4} + 2\,cos \,\frac{\pi}{4} = 2\,\sqrt{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.