What is the integral from 0 to 3 of xsqrt(4-x) ?

The integral from 0 to 3 of xsqrt(4-x) is 94/15

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integral from 0 to 3 of xsqrt(4-x)

\int_{0}^{3} x 4 - x d x

\frac{94}{15}

Decimal

6.26666 \dots

Solution steps

\int_{0}^{3} x 4 - x d x

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Expand

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

= - (- \int_{1}^{4} - 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

= - (- (- \int_{1}^{4} u^{\frac{3}{2}} d u + \int_{1}^{4} 4 u d u))

\int_{1}^{4} u^{\frac{3}{2}} d u = \frac{62}{5}

\int_{1}^{4} 4 u d u = \frac{56}{3}

= - (- (- \frac{62}{5} + \frac{56}{3}))

Simplify

- (- (- \frac{62}{5} + \frac{56}{3})) : \frac{94}{15}

= \frac{94}{15}

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

\int_{0}^{1} (10 x - 1)^{3} d x

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

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

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

What is the integral from 0 to 3 of xsqrt(4-x) ?
The integral from 0 to 3 of xsqrt(4-x) is 94/15