What is the integral from 3 to 7 of (x-6)sqrt(x-3) ?

The integral from 3 to 7 of (x-6)sqrt(x-3) is -16/5

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integral from 3 to 7 of (x-6)sqrt(x-3)

\int_{3}^{7} (x - 6) x - 3 d x

- \frac{16}{5}

Decimal

- 3.2

Solution steps

\int_{3}^{7} (x - 6) x - 3 d x

Apply Integration By Parts

: [\frac{2}{3} (x - 6) (x - 3)^{\frac{3}{2}} - \int \frac{2}{3} (x - 3)^{\frac{3}{2}} d x]_{3}^{7}

= [\frac{2}{3} (x - 6) (x - 3)^{\frac{3}{2}} - \int \frac{2}{3} (x - 3)^{\frac{3}{2}} d x]_{3}^{7}

\int \frac{2}{3} (x - 3)^{\frac{3}{2}} d x = \frac{4}{15} (x - 3)^{\frac{5}{2}}

= [\frac{2}{3} (x - 6) (x - 3)^{\frac{3}{2}} - \frac{4}{15} (x - 3)^{\frac{5}{2}}]_{3}^{7}

Simplify

[\frac{2}{3} (x - 6) (x - 3)^{\frac{3}{2}} - \frac{4}{15} (x - 3)^{\frac{5}{2}}]_{3}^{7} : [\frac{2}{5} (x - 8) (x - 3)^{\frac{3}{2}}]_{3}^{7}

= [\frac{2}{5} (x - 8) (x - 3)^{\frac{3}{2}}]_{3}^{7}

Compute the boundaries:

- \frac{16}{5}

= - \frac{16}{5}

\int_{1}^{e} 1

\int \frac{x ^{3}}{1 + x ^{2}} d x

x^{- \frac{1}{4}} + x^{\frac{1}{4}}

f (x) = x^{2} - 7, at x = 3

\frac{2}{x}, \frac{2}{x ^{2}}, 4

What is the integral from 3 to 7 of (x-6)sqrt(x-3) ?
The integral from 3 to 7 of (x-6)sqrt(x-3) is -16/5