What is the integral from 0 to 1 of (2x-3)(5x+1) ?

The integral from 0 to 1 of (2x-3)(5x+1) is -37/6

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integral from 0 to 1 of (2x-3)(5x+1)

\int_{0}^{1} (2 x - 3) (5 x + 1) d x

- \frac{37}{6}

Decimal Notation

- 6.16666 \dots

Solution steps

\int_{0}^{1} (2 x - 3) (5 x + 1) d x

Expand

(2 x - 3) (5 x + 1) : 10 x^{2} - 13 x - 3

= \int_{0}^{1} 10 x^{2} - 13 x - 3 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}^{1} 10 x^{2} d x - \int_{0}^{1} 13 x d x - \int_{0}^{1} 3 d x

\int_{0}^{1} 10 x^{2} d x = \frac{10}{3}

\int_{0}^{1} 13 x d x = \frac{13}{2}

\int_{0}^{1} 3 d x = 3

= \frac{10}{3} - \frac{13}{2} - 3

Simplify

\frac{10}{3} - \frac{13}{2} - 3 : - \frac{37}{6}

= - \frac{37}{6}

\int \frac{1}{2} sin (x) d x

x \to 8 + lim (lo g_{5} (x - 8))

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

\frac{d}{d x} (lo g_{5} (\frac{6 x ^{3} + 4}{2 x ^{3} - 3}))

(4, - 3), (12, 0)

What is the integral from 0 to 1 of (2x-3)(5x+1) ?
The integral from 0 to 1 of (2x-3)(5x+1) is -37/6