What is the integral from 0 to 4 of sqrt(16-y^2) ?

The integral from 0 to 4 of sqrt(16-y^2) is 4pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of sqrt(16-y^2)

\int_{0}^{4} 16 - y^{2} d y

4 π

Decimal Notation

12.56637 \dots

Solution steps

\int_{0}^{4} 16 - y^{2} d y

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 16 cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 16 cos^{2} (u) d u

Take the constant out:

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

= 16 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 16 \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

= 16 \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

= 16 \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

= 16 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

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

= 4 π

\int_{0}^{\frac{π}{3}} sin (2 x) sin (x) d x

\int_{8}^{15} \frac{1}{( x - 8 ) ^{7}} d x

\int_{3}^{s i n (x)} x^{3} d x

\int_{- \infty}^{0} 7^{r} d r

\int_{0}^{4} 9 + x^{2} d x

What is the integral from 0 to 4 of sqrt(16-y^2) ?
The integral from 0 to 4 of sqrt(16-y^2) is 4pi