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Chapter 3: Problem 1

Find the area enclosed between \(\mathrm{y}=\sin \mathrm{x}\) and \(x\)-axis as \(x\) varies from 0 to \(\frac{3 \pi}{2}\).

Short Answer

Expert verified
Answer: The area enclosed is 3 square units.

Step by step solution

01

Identifying the intersection points with the x-axis

To find the intersection points of the curve \(\mathrm{y}=\sin \mathrm{x}\) with the x-axis, we need to find the points for which \(\mathrm{y}=0\). This gives us the following equation: \(\sin \mathrm{x}=0\) In the given interval [0, \(\frac{3\pi}{2}\)], the solutions to this equation are 0, \(\pi\) and \(2\pi\). We will integrate the curve between these points and sum the integrals' absolute values to find the area enclosed between the curve and the x-axis.
02

Set up the integrals

We will set up three different integrals for each section between the intersection points as follows: 1. First section: from 0 to \(\pi\) \(\int_{0}^{\pi} \sin x\, dx\) 2. Second section: from \(\pi\) to \(2\pi\) \(\int_{\pi}^{2\pi} \sin x\, dx\)
03

Calculate the definite integrals

We will now compute the definite integrals and sum their absolute values to find the total area. 1. First section: \(\int_{0}^{\pi} \sin x\, dx = \left[-\cos x\right]_{0}^{\pi} = (-\cos \pi) - (-\cos 0) = 2\) 2. Second section: \(\int_{\pi}^{2\pi} \sin x\, dx = \left[-\cos x\right]_{\pi}^{2\pi} = (-\cos 2\pi) - (-\cos \pi) = 1\) The total area enclosed between the curve and the x-axis is the sum of the absolute values of these integrals: \(A = |2| + |1| = 2 +1 = 3\) Thus, the area enclosed between \(\mathrm{y}=\sin \mathrm{x}\) and the \(x\)-axis as \(x\) varies from 0 to \(\frac{3 \pi}{2}\) is 3 square units.

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Most popular questions from this chapter

Prove that the area common to the two ellipses \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) is \(4 a b \tan ^{-1} b / a\)

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