What is the integral from 0 to 7 of sqrt(-x^2+7x) ?

The integral from 0 to 7 of sqrt(-x^2+7x) is (49pi)/8

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integral from 0 to 7 of sqrt(-x^2+7x)

\int_{0}^{7} - x^{2} + 7 x d x

\frac{49 π}{8}

Decimal Notation

19.24225 \dots

Solution steps

\int_{0}^{7} - x^{2} + 7 x d x

Complete the square

- x^{2} + 7 x : - (x - \frac{7}{2})^{2} + \frac{49}{4}

= \int_{0}^{7} - (x - \frac{7}{2})^{2} + \frac{49}{4} d x

Apply u-substitution

: \int_{- \frac{7}{2}}^{\frac{7}{2}} \frac{- 4 u ^{2} + 49}{2} d u

= \int_{- \frac{7}{2}}^{\frac{7}{2}} \frac{- 4 u ^{2} + 49}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- \frac{7}{2}}^{\frac{7}{2}} - 4 u^{2} + 49 d u

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{49}{2} cos^{2} (v) d v

= \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{49}{2} cos^{2} (v) d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{49}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{49}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{49}{2} \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot \frac{49}{2} \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 v) d v)

\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v = π

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 v) d v = 0

= \frac{1}{2} \cdot \frac{49}{2} \cdot \frac{1}{2} (π + 0)

Simplify

\frac{1}{2} \cdot \frac{49}{2} \cdot \frac{1}{2} (π + 0) : \frac{49 π}{8}

= \frac{49 π}{8}

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What is the integral from 0 to 7 of sqrt(-x^2+7x) ?
The integral from 0 to 7 of sqrt(-x^2+7x) is (49pi)/8