What is the integral from-8 to 2 of sqrt(16-6x-x^2) ?

The integral from-8 to 2 of sqrt(16-6x-x^2) is (25pi)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-8 to 2 of sqrt(16-6x-x^2)

\int_{- 8}^{2} 16 - 6 x - x^{2} d x

\frac{25 π}{2}

Decimal

39.26990 \dots

Solution steps

\int_{- 8}^{2} 16 - 6 x - x^{2} d x

Complete the square

16 - 6 x - x^{2} : - (x + 3)^{2} + 25

= \int_{- 8}^{2} - (x + 3)^{2} + 25 d x

Apply u-substitution

: \int_{- 5}^{5} - u^{2} + 25 d u

= \int_{- 5}^{5} - u^{2} + 25 d u

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

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}}^{\frac{π}{2}} 1 + cos (2 v) d v

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}}^{\frac{π}{2}} 1 d v + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 v) d v)

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

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

= 25 \cdot \frac{1}{2} (π + 0)

Simplify

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

= \frac{25 π}{2}

\int_{0}^{\infty} ∣ sin (x) ∣ d x

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

\int_{0}^{1} 7 x (x^{2} - 1)^{5} d x

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

\int_{1}^{2} 3 x^{2} + 4 x d x

What is the integral from-8 to 2 of sqrt(16-6x-x^2) ?
The integral from-8 to 2 of sqrt(16-6x-x^2) is (25pi)/2