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Popular Calculus >

integral from-θ to θ of 1/(θ^2)(θ-|x|)

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Solution

∫−θθ​θ21​(θ−∣x∣)dx

Solution

2−θ∣θ∣​
Solution steps
∫−θθ​θ21​(θ−∣x∣)dx
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=θ21​⋅∫−θθ​θ−∣x∣dx
Apply the Sum Rule: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=θ21​(∫−θθ​θdx−∫−θθ​∣x∣dx)
∫−θθ​θdx=2θ2
∫−θθ​∣x∣dx=θ∣θ∣
=θ21​(2θ2−θ∣θ∣)
Simplify θ21​(2θ2−θ∣θ∣):2−θ∣θ∣​
=2−θ∣θ∣​

Popular Examples

integral from 0 to 1 of x^230x^2(1-x)^2∫01​x230x2(1−x)2dxintegral from 3 to infinity of 2/(x^2+9)∫3∞​x2+92​dxintegral from 0 to 1 of-4x+4∫01​−4x+4dxintegral from 0 to pi/2 of (sin(2x))^2∫02π​​(sin(2x))2dxintegral from-1 to 3 of cos(sqrt(x+1))∫−13​cos(x+1​)dx

Frequently Asked Questions (FAQ)

  • What is the integral from-θ to θ of 1/(θ^2)(θ-|x|) ?

    The integral from-θ to θ of 1/(θ^2)(θ-|x|) is 2-(|θ|)/θ
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