What is the integral from 0 to 2 of (x-3)^2-(x-1)^2 ?

The integral from 0 to 2 of (x-3)^2-(x-1)^2 is 8

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integral from 0 to 2 of (x-3)^2-(x-1)^2

\int_{0}^{2} (x - 3)^{2} - (x - 1)^{2} d x

8

Solution steps

\int_{0}^{2} (x - 3)^{2} - (x - 1)^{2} d x

Simplify

(x - 3)^{2} - (x - 1)^{2} : - 4 x + 8

= \int_{0}^{2} - 4 x + 8 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{2} 4 x d x + \int_{0}^{2} 8 d x

\int_{0}^{2} 4 x d x = 8

\int_{0}^{2} 8 d x = 16

= - 8 + 16

Simplify

= 8

(x + 1) y^{^{'}} + x y = 0

\frac{d}{d x} (\frac{2 x ^{2} - 1}{3 x ^{3} - 1})

\int e^{- x} sin^{2} (x) d x

\frac{d x}{d t} = \frac{4 x}{t}, x (1) = 1

\int \frac{e ^{2 x} ( 1 + 2 x ln ( x ) )}{4 x} d x

What is the integral from 0 to 2 of (x-3)^2-(x-1)^2 ?
The integral from 0 to 2 of (x-3)^2-(x-1)^2 is 8