High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Calculus 1 Textbook > Integrals

Computing Definite Integrals Geometrically

Properties of Definite IntegralsPrevious Section
The Fundamental Theorem of Calculus - Part 1Next Section
Popular Courses
Find My Course
0:00 / 0:00

Computing Definite Integrals Geometrically

Definite integrals represent area under a curve and can be computationally intensive. For certain functions, we can compute definite integrals simply by finding the area of common geometric shapes.

Geometric Integrals

The area under the graph of a continuous function f(x)f\left(x\right) on an interval [a,b]\left[a,b\right] is equal to ∫abf(x)dx\displaystyle \int_a^b f(x)dx.
Recall
  • Area of a Rectange=length×width\text{Area of a Rectange}=\text{length}\times \text{width}
  • Area of a Triangle=12×base×height\text{Area of a Triangle}=\frac{1}{2}\times \text{base} \times \text{height}
  • Area of a Circle=π×r2\text{Area of a Circle}=\pi\times r^2

Watch Out!
You may have to compute some integrals geometrically if you don't yet have the techniques of integration to do otherwise.

0:00 / 0:00

Example: Computing Integrals Geometrically

Compute the definite integral geometrically

∫−22(2−∣x∣)dx\displaystyle \int_{-2}^{2}(2-\lvert x\rvert)dx

∫−22(2−∣x∣)dx= area under the curve from x=−2 to x=2\displaystyle \int_{-2}^{2}(2-\lvert x\rvert)dx= \text{ area under the curve from } x=-2 \text{ to } x=2



∫−22(2−∣x∣)dx= area of the triangle\displaystyle \int_{-2}^{2}(2-\lvert x\rvert)dx= \text{ area of the triangle}

=12× base× height\displaystyle =\frac{1}{2} \times \text{ base} \times \text{ height}

=12×4×2\displaystyle =\frac{1}{2}\times4\times2

=4=\boxed4
Compute the following integral geometrically

∫0525−x2 dx\displaystyle \int_0^5 \sqrt{25-x^2} \ dx
Extra Practice
Practice: Circle
Q:\textbf{Q:} Find ∫244−(x−2)2\displaystyle\int_2^4\sqrt{4-\left(x-2\right)^2}
Definite Integrals
Calculate
∫01∣x−1∣dx\int_0^1|x-1|dx
(Hint: Sketch a Graph)
Practice: Circle
Q:\textbf{Q:} Find ∫244−(x−2)2\displaystyle\int_2^4\sqrt{4-\left(x-2\right)^2}
Integrals: Geometric Definitions
Evaluate the definite integral ∫5−65+6−x2+10x−19  ⁣dx\displaystyle\int_{5-\sqrt{6}}^{5+\sqrt{6}}\sqrt{-x^2+10x-19}\de{x}.
Evaluate ∫−111−x2dx\int_{-1}^1\sqrt{1-x^2}dx.
Hint: x2+y2=r2x^2+y^2=r^2 is a circle of radius r centered at (0,0)(0,0). Try to draw the curve in this integral.
Integrating Even and Odd Functions: Definite Integrals
Evaluate ∫−114−(x+1)2+sin⁡x⋅e2x4 dx\int_{-1}^1\sqrt{4-\left(x+1\right)^2}+\sin x\cdot e^{2x^4}\ dx.