What is the integral from 0 to 10 of (sqrt(100-x^2)) ?

The integral from 0 to 10 of (sqrt(100-x^2)) is 25pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 10 of (sqrt(100-x^2))

\int_{0}^{10} (100 - x^{2}) d x

25 π

Solution steps

\int_{0}^{10} 100 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 100 cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 100 cos^{2} (u) d u

Take the constant out:

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

= 100 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 100 \cdot \int_{0}^{\frac{π}{2}} \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

= 100 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 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

= 100 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

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

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

= 100 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

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

= 25 π

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

\int_{- 3}^{3} 5 d x

\int_{0}^{2} (3 x - 1)^{54} d x

\int_{1}^{9} 3 d x

\int_{3}^{6} \frac{12 x}{x ^{2} + 2 x - 8} d x

What is the integral from 0 to 10 of (sqrt(100-x^2)) ?
The integral from 0 to 10 of (sqrt(100-x^2)) is 25pi