What is the integral from 5 to 5sqrt(3 of)sqrt(100-x^2) ?

The integral from 5 to 5sqrt(3 of)sqrt(100-x^2) is (25pi)/3

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integral from 5 to 5sqrt(3 of)sqrt(100-x^2)

\int_{5}^{53} 100 - x^{2} d x

\frac{25 π}{3}

Decimal

26.17993 \dots

Solution steps

\int_{5}^{53} 100 - x^{2} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{3}} 100 cos^{2} (u) d u

= \int_{\frac{π}{6}}^{\frac{π}{3}} 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}{6}}^{\frac{π}{3}} cos^{2} (u) d u

Rewrite using trig identities

= 100 \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} \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}{6}}^{\frac{π}{3}} 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}{6}}^{\frac{π}{3}} 1 d u + \int_{\frac{π}{6}}^{\frac{π}{3}} cos (2 u) d u)

\int_{\frac{π}{6}}^{\frac{π}{3}} 1 d u = \frac{π}{6}

\int_{\frac{π}{6}}^{\frac{π}{3}} cos (2 u) d u = 0

= 100 \cdot \frac{1}{2} (\frac{π}{6} + 0)

Simplify

100 \cdot \frac{1}{2} (\frac{π}{6} + 0) : \frac{25 π}{3}

= \frac{25 π}{3}

\int e^{\frac{1}{2}}

cos^{2} (x) - sin^{2} (x)

y^{^{'}} + \frac{3 y}{x} = \frac{2}{x}

\int_{0}^{π} 15 sin^{2} (2 x) d x

f (x) = - \frac{1}{2} x^{2} - 2 x + 3, at x = 0

What is the integral from 5 to 5sqrt(3 of)sqrt(100-x^2) ?
The integral from 5 to 5sqrt(3 of)sqrt(100-x^2) is (25pi)/3