What is the integral from 5 to 1 of x^2sqrt(x-1) ?

The integral from 5 to 1 of x^2sqrt(x-1) is -7088/105

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integral from 5 to 1 of x^2sqrt(x-1)

\int_{5}^{1} x^{2} x - 1 d x

- \frac{7088}{105}

Decimal

- 67.50476 \dots

Solution steps

\int_{5}^{1} x^{2} x - 1 d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{5} x^{2} x - 1 d x

Apply u-substitution

: \int_{0}^{4} (u + 1)^{2} u d u

= - \int_{0}^{4} (u + 1)^{2} u d u

Expand

(u + 1)^{2} u : u^{\frac{5}{2}} + 2 u^{\frac{3}{2}} + u

= - \int_{0}^{4} u^{\frac{5}{2}} + 2 u^{\frac{3}{2}} + u d u

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} u^{\frac{5}{2}} d u + \int_{0}^{4} 2 u^{\frac{3}{2}} d u + \int_{0}^{4} u d u)

\int_{0}^{4} u^{\frac{5}{2}} d u = \frac{256}{7}

\int_{0}^{4} 2 u^{\frac{3}{2}} d u = \frac{128}{5}

\int_{0}^{4} u d u = \frac{16}{3}

= - (\frac{256}{7} + \frac{128}{5} + \frac{16}{3})

Join

\frac{256}{7} + \frac{128}{5} + \frac{16}{3} : \frac{7088}{105}

= - \frac{7088}{105}

\int_{- 6}^{2} x - 12 - 4 x d x

\int_{0.5}^{0.75} 12 x^{2} d x

\int_{1}^{3} \frac{3}{7 x - 2} d x

\int_{x}^{4} \frac{6 ( y - x )}{64} d y

\int_{0}^{l o g_{6} (2)} 6^{x} d x

What is the integral from 5 to 1 of x^2sqrt(x-1) ?
The integral from 5 to 1 of x^2sqrt(x-1) is -7088/105