What is the integral from-9 to 0 of sqrt(81-x^2) ?

The integral from-9 to 0 of sqrt(81-x^2) is (81pi)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from-9 to 0 of sqrt(81-x^2)

\int_{- 9}^{0} 81 - x^{2} d x

\frac{81 π}{4}

Decimal

63.61725 \dots

Solution steps

\int_{- 9}^{0} 81 - x^{2} d x

Apply Trigonometric Substitution

= \int_{- \frac{π}{2}}^{0} 81 cos^{2} (u) d u

Take the constant out:

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

= 81 \cdot \int_{- \frac{π}{2}}^{0} cos^{2} (u) d u

Rewrite using trig identities

= 81 \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

= 81 \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

= 81 \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

= 81 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

81 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{81 π}{4}

= \frac{81 π}{4}

\int_{3}^{4} x^{2} - 9 d x

\int_{0}^{π} x sin (\frac{n x}{2}) d x

\int_{0}^{1} x^{2} cos (nπ x) d x

\int_{0}^{1} x^{3} 1 + x^{2} d x

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

What is the integral from-9 to 0 of sqrt(81-x^2) ?
The integral from-9 to 0 of sqrt(81-x^2) is (81pi)/4