Evaluate the integral and interpret it as the area of a region. π/2 15 sin x 5 cos 2x dx Sketch the region. O y 4 2 X 6 -4 MA NA 4 y 2 y O 4 4 2 -2 X X
Evaluate the integral and interpret it as the area of a region. π/2 15 sin x 5 cos 2x dx Sketch the region. O y 4 2 X 6 -4 MA NA 4 y 2 y O 4 4 2 -2 X X
Evaluate the integral and interpret it as the area of a region. π/2 15 sin x 5 cos 2x dx Sketch the region. O y 4 2 X 6 -4 MA NA 4 y 2 y O 4 4 2 -2 X X
Evaluate the integral and interpret it as the area of a region.\
Could you please explain each curve and how it was obtained as well.
Transcribed Image Text:**Evaluate the Integral and Interpret it as the Area of a Region**
Evaluate the integral:
\[
\int_{0}^{\pi/2} |5 \sin x - 5 \cos 2x| \, dx
\]
**Sketch the Region**
Four graphs are presented with shaded areas representing the regions related to the integral given.
1. **Top Left Graph:**
- The graph shows a sinusoidal curve oscillating between \( y = 4 \) and \( y = -4 \).
- The region of interest is a symmetric shaded area that intersects the x-axis at \( x = \frac{\pi}{6} \).
- The region is bounded by \( x = 0 \) and \( x = \frac{\pi}{2} \).
2. **Top Right Graph:**
- This graph features a different sinusoidal section with positive area above the x-axis.
- The shaded region starts at the origin and peaks before \( x = \frac{\pi}{2} \).
3. **Bottom Left Graph:**
- This illustration depicts two symmetrical lobes with a shaded area above the x-axis initially, followed by a negative lobe below.
- The positive portion is shaded, extending from \( x = 0 \) to \( x = \frac{\pi/2} \).
4. **Bottom Right Graph:**
- The graph depicts a continuous curve with an entirely positive shaded area from the origin to \( x = \frac{\pi}{2} \).
- This indicates the entire interval from \( x = 0 \) to \( x = \frac{\pi/2} \) contributes positively.
These graphs illustrate potential interpretations of the function's behavior under the integral, emphasizing its sinusoidal nature and varied regions that contribute to the integral's total area.
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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![**Evaluate the Integral and Interpret it as the Area of a Region**
Evaluate the integral:
\[
\int_{0}^{\pi/2} |5 \sin x - 5 \cos 2x| \, dx
\]
**Sketch the Region**
Four graphs are presented with shaded areas representing the regions related to the integral given.
1. **Top Left Graph:**
- The graph shows a sinusoidal curve oscillating between \( y = 4 \) and \( y = -4 \).
- The region of interest is a symmetric shaded area that intersects the x-axis at \( x = \frac{\pi}{6} \).
- The region is bounded by \( x = 0 \) and \( x = \frac{\pi}{2} \).
2. **Top Right Graph:**
- This graph features a different sinusoidal section with positive area above the x-axis.
- The shaded region starts at the origin and peaks before \( x = \frac{\pi}{2} \).
3. **Bottom Left Graph:**
- This illustration depicts two symmetrical lobes with a shaded area above the x-axis initially, followed by a negative lobe below.
- The positive portion is shaded, extending from \( x = 0 \) to \( x = \frac{\pi/2} \).
4. **Bottom Right Graph:**
- The graph depicts a continuous curve with an entirely positive shaded area from the origin to \( x = \frac{\pi}{2} \).
- This indicates the entire interval from \( x = 0 \) to \( x = \frac{\pi/2} \) contributes positively.
These graphs illustrate potential interpretations of the function's behavior under the integral, emphasizing its sinusoidal nature and varied regions that contribute to the integral's total area.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fea000c69-b4f2-413a-9977-6a5d2d2ee8af%2F7a6854a5-5594-4a80-bdfd-f61bc9c482c8%2Frgpd66_processed.jpeg&w=3840&q=75)
