EXAMPLE 5 Find the area of the region bounded by the curves y = sin(x), y = cos(x), x = 0, and x = 7/2. SOLUTION The points of intersection occur when sin(x) = cos(x), that is, when x = (since 0sxs n/2). The region is sketched in the figure. Observe that cos(x) > sin(x) when , but sin(x) 2 cos(x) when Therefore the required area is 피2 ! Icos(x) – sin(x)| dx = A1 + A2 A = 7/4 /2 | (cos(x) – sin(x)) dx + 7/4 (sin(x) – cos(x)) dx 7/4 7/2 + 피/4 -0 - 1 + -0 - 1 + = 2/2 - 2. In this particular example we could have saved some work by noticing that the region is symmetric about x = n/4 and so /4 A = 2A, = 2 (cos(x) – sin(x)) dx. Jo

Calculus: Early Transcendentals
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Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Question
**Example 5**

Find the area of the region bounded by the curves \( y = \sin(x) \), \( y = \cos(x) \), \( x = 0 \), and \( x = \pi/2 \).

**Solution**

The points of intersection occur when \( \sin(x) = \cos(x) \), that is, when \( x = \pi/4 \) (since \( 0 \leq x \leq \pi/2 \)). The region is sketched in the figure. Observe that \( \cos(x) \geq \sin(x) \) when \( 0 \leq x \leq \pi/4 \), but \( \sin(x) \geq \cos(x) \) when \( \pi/4 \leq x \leq \pi/2 \). Therefore, the required area is

\[
A = \int_{0}^{\pi/2} |\cos(x) - \sin(x)| \, dx = A_1 + A_2
\]

\[
= \int_{0}^{\pi/4} (\cos(x) - \sin(x)) \, dx + \int_{\pi/4}^{\pi/2} (\sin(x) - \cos(x)) \, dx
\]

\[
= \left[ \boxed{0 - 1} \right]_{0}^{\pi/4} + \left[ \boxed{-0 - 1 +} \right]_{\pi/4}^{\pi/2}
\]

\[
= 2\sqrt{2} - 2
\]

In this particular example, we could have saved some work by noticing that the region is symmetric about \( x = \pi/4 \) and so

\[
A = 2A_1 = 2 \int_{0}^{\pi/4} (\cos(x) - \sin(x)) \, dx
\]
Transcribed Image Text:**Example 5** Find the area of the region bounded by the curves \( y = \sin(x) \), \( y = \cos(x) \), \( x = 0 \), and \( x = \pi/2 \). **Solution** The points of intersection occur when \( \sin(x) = \cos(x) \), that is, when \( x = \pi/4 \) (since \( 0 \leq x \leq \pi/2 \)). The region is sketched in the figure. Observe that \( \cos(x) \geq \sin(x) \) when \( 0 \leq x \leq \pi/4 \), but \( \sin(x) \geq \cos(x) \) when \( \pi/4 \leq x \leq \pi/2 \). Therefore, the required area is \[ A = \int_{0}^{\pi/2} |\cos(x) - \sin(x)| \, dx = A_1 + A_2 \] \[ = \int_{0}^{\pi/4} (\cos(x) - \sin(x)) \, dx + \int_{\pi/4}^{\pi/2} (\sin(x) - \cos(x)) \, dx \] \[ = \left[ \boxed{0 - 1} \right]_{0}^{\pi/4} + \left[ \boxed{-0 - 1 +} \right]_{\pi/4}^{\pi/2} \] \[ = 2\sqrt{2} - 2 \] In this particular example, we could have saved some work by noticing that the region is symmetric about \( x = \pi/4 \) and so \[ A = 2A_1 = 2 \int_{0}^{\pi/4} (\cos(x) - \sin(x)) \, dx \]
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