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

The integral from 4 to 5 of 3xsqrt(x-4) is 46/5

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

\int_{4}^{5} 3 x x - 4 d x

\frac{46}{5}

Decimal Notation

9.2

Solution steps

\int_{4}^{5} 3 x x - 4 d x

Take the constant out:

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

= 3 \cdot \int_{4}^{5} x x - 4 d x

Apply u-substitution

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

= 3 \cdot \int_{0}^{1} (u + 4) u d u

Expand

(u + 4) u : u^{\frac{3}{2}} + 4 u

= 3 \cdot \int_{0}^{1} u^{\frac{3}{2}} + 4 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}^{1} u^{\frac{3}{2}} d u + \int_{0}^{1} 4 u d u)

\int_{0}^{1} u^{\frac{3}{2}} d u = \frac{2}{5}

\int_{0}^{1} 4 u d u = \frac{8}{3}

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

Simplify

3 (\frac{2}{5} + \frac{8}{3}) : \frac{46}{5}

= \frac{46}{5}

\int_{1}^{4} (2 x + 3) d x

\int_{- 1}^{0} 1 - x^{2} d x

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

\int_{0}^{3} \frac{x ^{2}}{9} d x

\int_{- π}^{0} - 1 d x

What is the integral from 4 to 5 of 3xsqrt(x-4) ?
The integral from 4 to 5 of 3xsqrt(x-4) is 46/5