What is the integral from 7 to 14 of sqrt(196-x^2) ?

The integral from 7 to 14 of sqrt(196-x^2) is 98(pi/3-(sqrt(3))/4)

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integral from 7 to 14 of sqrt(196-x^2)

\int_{7}^{14} 196 - x^{2} d x

98 (\frac{π}{3} - \frac{3}{4})

Decimal Notation

60.19011 \dots

Solution steps

\int_{7}^{14} 196 - x^{2} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 196 cos^{2} (u) d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 196 cos^{2} (u) d u

Take the constant out:

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

= 196 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 196 \cdot \int_{\frac{π}{6}}^{\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

= 196 \cdot \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\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

= 196 \cdot \frac{1}{2} (\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u + \int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 u) d u)

\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u = \frac{π}{3}

\int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 u) d u = - \frac{3}{4}

= 196 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

Simplify

196 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : 98 (\frac{π}{3} - \frac{3}{4})

= 98 (\frac{π}{3} - \frac{3}{4})

\int \frac{1}{3} e^{x + 3} + \frac{3}{x} - 5 d x

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

\frac{2 t}{2 + t}

t (\frac{d y}{d t}) + 34 y = \frac{1}{t}

y = 3^{c o s (x)}

What is the integral from 7 to 14 of sqrt(196-x^2) ?
The integral from 7 to 14 of sqrt(196-x^2) is 98(pi/3-(sqrt(3))/4)