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

The integral from-4 to 4 of (sqrt(16-x^2)) is 8pi

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

\int_{- 4}^{4} (16 - x^{2}) d x

8 π

Decimal Notation

25.13274 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= 16 \cdot \int_{- \frac{π}{2}}^{\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_{- \frac{π}{2}}^{\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_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 u) d u)

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

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

= 16 \cdot \frac{1}{2} (π + 0)

Simplify

16 \cdot \frac{1}{2} (π + 0) : 8 π

= 8 π

y^{^{'}} = \frac{x}{x - 1}, y (2) = 5

\int \frac{x}{x ^{2} - 8 x} d x

\frac{d}{d x} (5 x sin (x))

\frac{d}{d x} (3 x^{2} - 2 x + 4)

What is the integral from-4 to 4 of (sqrt(16-x^2)) ?
The integral from-4 to 4 of (sqrt(16-x^2)) is 8pi