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

The integral from-10 to 10 of sqrt(100-x^2) is 50pi

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

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

50 π

Decimal

157.07963 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= 100 \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

= 100 \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

= 100 \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

= 100 \cdot \frac{1}{2} (π + 0)

Simplify

100 \cdot \frac{1}{2} (π + 0) : 50 π

= 50 π

cos (y) y^{^{'}} + \frac{1}{x} sin (y) = 1

y = 2 x, y = 2, x = 0

50

y^{^{'}} = \frac{y ^{3}}{e ^{x}}, y (0) = - 1

\frac{d y}{d x} = (4 - y)^{5}

What is the integral from-10 to 10 of sqrt(100-x^2) ?
The integral from-10 to 10 of sqrt(100-x^2) is 50pi