What is the integral from 0 to 5 of (x-4)^2 ?

The integral from 0 to 5 of (x-4)^2 is 65/3

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integral from 0 to 5 of (x-4)^2

\int_{0}^{5} (x - 4)^{2} d x

\frac{65}{3}

Decimal

21.66666 \dots

Solution steps

\int_{0}^{5} (x - 4)^{2} d x

Expand

(x - 4)^{2} : x^{2} - 8 x + 16

= \int_{0}^{5} x^{2} - 8 x + 16 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}^{5} x^{2} d x - \int_{0}^{5} 8 x d x + \int_{0}^{5} 16 d x

\int_{0}^{5} x^{2} d x = \frac{125}{3}

\int_{0}^{5} 8 x d x = 100

\int_{0}^{5} 16 d x = 80

= \frac{125}{3} - 100 + 80

Simplify

\frac{125}{3} - 100 + 80 : \frac{65}{3}

= \frac{65}{3}

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\int_{0}^{π} 8 cos^{2} (x) d x

\int_{0}^{2} \frac{1}{2 x - 1} d x

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\int_{1}^{5} \frac{2 x}{( x ^{2} + 1 ) ^{2}} d x

What is the integral from 0 to 5 of (x-4)^2 ?
The integral from 0 to 5 of (x-4)^2 is 65/3