The value of \(∫_{\frac \pi6}^{\frac \pi3}\sqrt {1-sin2x\ dx}\) is
\(\sqrt 2-\sqrt 3+1\)
\(2\sqrt 2-\sqrt 3-1\)
\(2\sqrt 2+\sqrt 3-1\)
\(\sqrt 2+\sqrt 3-1\)
The Correct Option is B
Solution and Explanation
The correct option is (B): \(2\sqrt 2-\sqrt 3-1\).
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Questions Asked in JEE Main exam
- The area of the region enclosed by the curves \( y = x^2 - 4x + 4 \) and \( y^2 = 16 - 8x \) is:
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- Sequences and Series
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).
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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.