The area (in square units) of the region bounded by the curve y = |sin2x| and the X-axis in [0,2π] is
The area (in square units) of the region bounded by the curve y = |sin2x| and the X-axis in [0,2π] is
0
3
4
1
The Correct Option is C
Solution and Explanation
To solve the problem, we need to find the area under the curve $y = |\sin 2x|$ over the interval $[0, 2\pi]$.
1. Set Up the Integral:
Since $y = |\sin 2x|$ is non-negative, the area is:
$\int_0^{2\pi} |\sin 2x| dx$.
2. Use Periodicity:
The period of $\sin 2x$ is $\pi$, so $|\sin 2x|$ has period $\pi/2$.
Thus, the integral over $[0, 2\pi]$ (four periods) is:
$\int_0^{2\pi} |\sin 2x| dx = 4 \int_0^{\pi/2} |\sin 2x| dx$.
Since $\sin 2x \ge 0$ on $[0, \pi/2]$, $|\sin 2x| = \sin 2x$ there.
3. Evaluate the Integral:
Compute:
$4 \int_0^{\pi/2} \sin 2x dx = 4 \left[ -\frac{\cos 2x}{2} \right]_0^{\pi/2} = 4 \left( -\frac{\cos \pi}{2} - \left( -\frac{\cos 0}{2} \right) \right)$.
Simplify:
$= 4 \left( \frac{1}{2} + \frac{1}{2} \right) = 4 \cdot 1 = 4$.
Final Answer:
The area is $4$.
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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.