What is the integral from 2 to 7 of (x^2-2x)^2 ?

The integral from 2 to 7 of (x^2-2x)^2 is 4250/3

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integral from 2 to 7 of (x^2-2x)^2

\int_{2}^{7} (x^{2} - 2 x)^{2} d x

\frac{4250}{3}

Decimal Notation

1416.66666 \dots

Solution steps

\int_{2}^{7} (x^{2} - 2 x)^{2} d x

Expand

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

= \int_{2}^{7} x^{4} - 4 x^{3} + 4 x^{2} 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_{2}^{7} x^{4} d x - \int_{2}^{7} 4 x^{3} d x + \int_{2}^{7} 4 x^{2} d x

\int_{2}^{7} x^{4} d x = 3355

\int_{2}^{7} 4 x^{3} d x = 2385

\int_{2}^{7} 4 x^{2} d x = \frac{1340}{3}

= 3355 - 2385 + \frac{1340}{3}

Simplify

3355 - 2385 + \frac{1340}{3} : \frac{4250}{3}

= \frac{4250}{3}

\int \frac{( x ^{3} - 2 x ^{2} + 7 x - 7 )}{x ^{2} - 2 x + 5} d x

f (x) = - e^{- x}

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

f (x) = x^{2} + 4, g (x) = - x^{2} - 1, [0, 1]

f (x) = 6 x^{2} + \frac{4}{x ^{2}} - 10, at x = 1

What is the integral from 2 to 7 of (x^2-2x)^2 ?
The integral from 2 to 7 of (x^2-2x)^2 is 4250/3