What is the integral from 1 to 4 of sqrt(25-5x) ?

The integral from 1 to 4 of sqrt(25-5x) is (14sqrt(5))/3

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integral from 1 to 4 of sqrt(25-5x)

\int_{1}^{4} 25 - 5 x d x

\frac{14 5}{3}

Decimal Notation

10.43498 \dots

Solution steps

\int_{1}^{4} 25 - 5 x d x

Apply u-substitution

: \int_{20}^{5} - \frac{u}{5} d u

= \int_{20}^{5} - \frac{u}{5} d u

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

= - \int_{5}^{20} - \frac{u}{5} d u

Take the constant out:

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

= - (- \frac{1}{5} \cdot \int_{5}^{20} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{20}

= - (- \frac{1}{5} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{20})

Simplify

= \frac{1}{5} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{20}

Compute the boundaries:

\frac{70 5}{3}

= \frac{1}{5} \cdot \frac{70 5}{3}

Simplify

= \frac{14 5}{3}

\int_{0}^{6 - 2 x} (6 - 2 x - y) d y

\int_{- 0}^{+ 5} \frac{x}{4 + x ^{2}} d x

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

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

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

What is the integral from 1 to 4 of sqrt(25-5x) ?
The integral from 1 to 4 of sqrt(25-5x) is (14sqrt(5))/3