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

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

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

4 π

Decimal Notation

12.56637 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= 16 \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

= 16 \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

= 16 \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

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

Simplify

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

= 4 π

\frac{d y}{d x} = (x + 1)^{3}

\int (x + 1) x + 6 d x

\int 3 sec^{6} (t) d t

\frac{\partial}{\partial x} (3 cos^{2} (x))

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