Wize AP Calculus (BC) Textbook > Integration & Accumulation of Change

Computing Definite Integrals Geometrically

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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)f(x) on an interval [a,b]\left[a,b\right][a,b] is equal to ∫abf(x)dx\displaystyle \int_a^b f(x)dx∫ab​f(x)dx. 
Recall
Area of a Rectange=length×width\text{Area of a Rectange}=\text{length}\times \text{width}Area of a Rectange=length×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 Triangle=21​×base×height
Area of a Circle=π×r2\text{Area of a Circle}=\pi\times r^2Area of a Circle=π×r2

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

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 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

∫−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 under the curve from x=−2 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}∫−22​(2−∣x∣)dx= area of the triangle

=12× base× height\displaystyle =\frac{1}{2} \times \text{ base} \times \text{ height}=21​× base× height

=12×4×2\displaystyle =\frac{1}{2}\times4\times2=21​×4×2

=4=\boxed4=4​

Compute the following integral geometrically

∫0525−x2 dx\displaystyle \int_0^5 \sqrt{25-x^2} \ dx∫05​25−x2​ dx

Enter the exact value of the definite integral.

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