NEW! Section 6.1 – Area
Instructions
- The first videos below explain the concepts in this section.
- This page also includes exercises that you should attempt to solve yourself. You can check your answers and watch the videos explaining how to solve the exercises.
- You can find additional practice problems for this section on the Practice Problems page for this Module.
Concepts
- Finding the area between two curves on a closed interval
- Finding the area bounded by two curves
- Determining whether to integrate along the x or y-axis when finding the area between curves
Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Sketch the region \(R\) bounded by \(y=\sin x\), \(y=0\), \(x=0\), \(x=\displaystyle{{\pi}\over{3}}\). Find the area of an arbitrary rectangle. What is the area of \(R?\)
Area of an arbitrary rectangle: \( A_i=\sin(x_i)\Delta x_i\)
Area \(=\displaystyle\int_{0}^{\pi/3} \sin x \, dx= \frac{1}{2}\)
See the video for the sketch of the region.
Area \(=\displaystyle\int_{0}^{\pi/3} \sin x \, dx= \frac{1}{2}\)
See the video for the sketch of the region.
2. Sketch the region \(R\) bounded by \(y=x^2+1\) and \(y=3-x^2\). Find the area of an arbitrary rectangle. What is the area of \(R?\)
Area of an arbitrary rectangle: \(A_i=\left( 2-2x_i^2\right)\Delta x_i\)
Area: \(A=\displaystyle \int_{-1}^1 \left(2-2x^2\right)\, dx=\frac{8}{3}\)
See the video for a sketch of the region.
Area: \(A=\displaystyle \int_{-1}^1 \left(2-2x^2\right)\, dx=\frac{8}{3}\)
See the video for a sketch of the region.
3. Sketch the region \(R\) bounded by bounded by \(x=2-2y^2\), \(x=2y^2-2\). Find the area of an arbitrary rectangle. What is the area of \(R?\)
Area of a rectangle: \(A_i=\left[ 4-4y_i^2\right]\Delta y_i\)
Area: \(A=\displaystyle \int_{y=-1}^{y=1} \left(4-4y^2\right)\, dy = \frac{16}{3}\)
See the video for a sketch of the region.
Area: \(A=\displaystyle \int_{y=-1}^{y=1} \left(4-4y^2\right)\, dy = \frac{16}{3}\)
See the video for a sketch of the region.
4. Sketch the region \(R\) bounded by \(y^2=x\), \(x-2y=3\). Set up but do not evaluate an integral in terms of \(x\), then \(y\), that gives the area of \(R.\)
Integral in terms of \(y\): \(A=\displaystyle \int_{-1}^3 \left(2y+3-y^2\right) \, dy\)
Integral in terms of \(x\): \(A=\displaystyle \int_0^1 \left[\sqrt{x}-\left(-\sqrt{x}\right)\right]\, dx + \int_{1}^{9} \left[ \sqrt{x}-\left( \frac{x-3}{2}\right)\right] \, dx\)
See the video for a sketch of the region.
Integral in terms of \(x\): \(A=\displaystyle \int_0^1 \left[\sqrt{x}-\left(-\sqrt{x}\right)\right]\, dx + \int_{1}^{9} \left[ \sqrt{x}-\left( \frac{x-3}{2}\right)\right] \, dx\)
See the video for a sketch of the region.
5. Sketch the region \(R\) bounded by \(y=\sin x\), \(y=\cos x\), \(x=-\displaystyle{{\pi}\over{2}}\), \(x=\displaystyle{{\pi}\over{2}}\). Find the area of \(R.\)
\(A=\displaystyle \int_{-\pi/2}^{\pi/4} (\cos x -\sin x)\, dx + \int_{\pi/4}^{\pi/2} (\sin x-\cos x) \, dx=2\sqrt{2}\)
See the video for a sketch of the region.
See the video for a sketch of the region.