What is the integral from 1 to 3 of 3xsqrt(x-1) ?

The integral from 1 to 3 of 3xsqrt(x-1) is (44sqrt(2))/5

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integral from 1 to 3 of 3xsqrt(x-1)

\int_{1}^{3} 3 x x - 1 d x

\frac{44 2}{5}

Decimal

12.44507 \dots

Solution steps

\int_{1}^{3} 3 x x - 1 d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{1}^{3} x x - 1 d x

Apply u-substitution

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

= 3 \cdot \int_{0}^{2} (u + 1) u d u

Expand

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

= 3 \cdot \int_{0}^{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

= 3 (\int_{0}^{2} u^{\frac{3}{2}} d u + \int_{0}^{2} u d u)

\int_{0}^{2} u^{\frac{3}{2}} d u = \frac{8 2}{5}

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

= 3 (\frac{8 2}{5} + \frac{4 2}{3})

Simplify

3 (\frac{8 2}{5} + \frac{4 2}{3}) : \frac{44 2}{5}

= \frac{44 2}{5}

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

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

\int_{1}^{2} (1) d x

\int_{- 4}^{6} (∣ x ∣) d x

\int_{2}^{3} (2 - x^{2}) d x

What is the integral from 1 to 3 of 3xsqrt(x-1) ?
The integral from 1 to 3 of 3xsqrt(x-1) is (44sqrt(2))/5