What is the integral from-5 to 0 of sqrt(25-x^2) ?

The integral from-5 to 0 of sqrt(25-x^2) is (25pi)/4

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integral from-5 to 0 of sqrt(25-x^2)

\int_{- 5}^{0} 25 - x^{2} d x

\frac{25 π}{4}

Decimal Notation

19.63495 \dots

Solution steps

\int_{- 5}^{0} 25 - x^{2} d x

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{0} 25 cos^{2} (u) d u

= \int_{- \frac{π}{2}}^{0} 25 cos^{2} (u) d u

Take the constant out:

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

= 25 \cdot \int_{- \frac{π}{2}}^{0} cos^{2} (u) d u

Rewrite using trig identities

= 25 \cdot \int_{- \frac{π}{2}}^{0} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= 25 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{0} 1 + cos (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

= 25 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{0} 1 d u + \int_{- \frac{π}{2}}^{0} cos (2 u) d u)

\int_{- \frac{π}{2}}^{0} 1 d u = \frac{π}{2}

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

= 25 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

25 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{25 π}{4}

= \frac{25 π}{4}

\frac{d}{d x} (\frac{4 x ^{2} + 8 x + 8}{x})

\frac{d}{d x} (arctan (\frac{x}{b}))

\int sec^{2} (\frac{x}{2}) d x

f (x) = - 2 x^{3}, at x = - 1

(1, - \frac{1}{2}), (- 2, - \frac{7}{2})

What is the integral from-5 to 0 of sqrt(25-x^2) ?
The integral from-5 to 0 of sqrt(25-x^2) is (25pi)/4