What is the integral from 0 to 4 of (x-3)(x-5) ?

The integral from 0 to 4 of (x-3)(x-5) is 52/3

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integral from 0 to 4 of (x-3)(x-5)

\int_{0}^{4} (x - 3) (x - 5) d x

\frac{52}{3}

Decimal Notation

17.33333 \dots

Solution steps

\int_{0}^{4} (x - 3) (x - 5) d x

Expand

(x - 3) (x - 5) : x^{2} - 8 x + 15

= \int_{0}^{4} x^{2} - 8 x + 15 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}^{4} x^{2} d x - \int_{0}^{4} 8 x d x + \int_{0}^{4} 15 d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 8 x d x = 64

\int_{0}^{4} 15 d x = 60

= \frac{64}{3} - 64 + 60

Simplify

\frac{64}{3} - 64 + 60 : \frac{52}{3}

= \frac{52}{3}

\int_{1}^{4} \frac{1}{x + 1} d x

\int_{1}^{4} 2 x + 1 d x

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

\int_{0}^{1} (4 x - 3 x^{3}) d x

\int_{0}^{\frac{π}{2}} cos (x) + sin (x) d x

What is the integral from 0 to 4 of (x-3)(x-5) ?
The integral from 0 to 4 of (x-3)(x-5) is 52/3