Applications of Integration

AP Calculus BC guide to integration applications: area, volume (disk, washer, shell), arc length, and cross-sectional volume methods.

# Applications of Integration — AP Calculus BC

Integration is used to calculate areas, volumes, and arc lengths. These applications are tested heavily on the AP Calculus BC exam, especially in the free-response section.

Key Concepts

Area Between Curves

A=abf(x)g(x)dxA = \int_a^b |f(x) - g(x)|\,dx or for curves described as functions of yy: A=cdf(y)g(y)dyA = \int_c^d |f(y) - g(y)|\,dy

Volume: Disk Method

Rotating y=f(x)y = f(x) around the xx-axis: V=πab[f(x)]2dxV = \pi\int_a^b [f(x)]^2\,dx

Volume: Washer Method

Rotating the region between f(x)f(x) and g(x)g(x) (f>g0f > g \geq 0): V=πab([f(x)]2[g(x)]2)dxV = \pi\int_a^b \left([f(x)]^2 - [g(x)]^2\right)dx

Volume: Shell Method

Rotating around the yy-axis: V=2πabxf(x)dxV = 2\pi\int_a^b x \cdot f(x)\,dx

Volume by Cross-Sections

If cross-sections perpendicular to the xx-axis have area A(x)A(x): V=abA(x)dxV = \int_a^b A(x)\,dx

Common cross-sections: squares, semicircles, equilateral triangles.

Arc Length

L=ab1+[f(x)]2dxL = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx

Worked Example

Problem: Find the volume when y=xy = \sqrt{x} from x=0x = 0 to x=4x = 4 is rotated about the xx-axis.

Solution: Disk method: V=π04(x)2dx=π04xdx=π[x22]04=π(8)=8πV = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi(8) = 8\pi

Practice Questions

  1. 1. Find the area between y=x2y = x^2 and y=xy = x from x=0x = 0 to x=1x = 1.

    A=01(xx2)dx=[x2/2x3/3]01=1/21/3=1/6A = \int_0^1 (x - x^2)\,dx = [x^2/2 - x^3/3]_0^1 = 1/2 - 1/3 = 1/6.

    2. Find the volume when the region between y=xy = x and y=x2y = x^2 (from 0 to 1) is rotated about the xx-axis.

    Washer: V=π01(x2x4)dx=π[x3/3x5/5]01=π(1/31/5)=2π/15V = \pi\int_0^1 (x^2 - x^4)\,dx = \pi[x^3/3 - x^5/5]_0^1 = \pi(1/3 - 1/5) = 2\pi/15.

    3. A solid has a base bounded by y=0y = 0 and y=1x2y = 1 - x^2. Cross-sections perpendicular to the xx-axis are squares. Find the volume.

    Side length =1x2= 1 - x^2. V=11(1x2)2dx=11(12x2+x4)dx=2[x2x3/3+x5/5]01=2(12/3+1/5)=2(8/15)=16/15V = \int_{-1}^{1}(1-x^2)^2\,dx = \int_{-1}^{1}(1 - 2x^2 + x^4)\,dx = 2[x - 2x^3/3 + x^5/5]_0^1 = 2(1 - 2/3 + 1/5) = 2(8/15) = 16/15.

Want to check your answers and get step-by-step solutions?

Get it on Google PlayDownload on the App Store

Summary

  • Area: fg\int |f - g|.
  • Volume of revolution: disk (πr2\pi r^2), washer (π(R2r2)\pi(R^2 - r^2)), shell (2πrh2\pi rh).
  • Volume by cross-sections: A(x)dx\int A(x)\,dx.
  • Arc length: 1+(f)2dx\int\sqrt{1 + (f')^2}\,dx.

Ready to Ace Your AP CALCULUS BC calculus?

Get instant step-by-step solutions to any problem. Snap a photo and learn with Tutor AI — your personal exam prep companion.

Get it on Google PlayDownload on the App Store