What is the integral from-7 to 7 of (sqrt(49-x^2)) ?

The integral from-7 to 7 of (sqrt(49-x^2)) is (49pi)/2

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integral from-7 to 7 of (sqrt(49-x^2))

\int_{- 7}^{7} (49 - x^{2}) d x

\frac{49 π}{2}

Decimal Notation

76.96902 \dots

Solution steps

\int_{- 7}^{7} 49 - x^{2} d x

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} 49 cos^{2} (u) d u

= \int_{- \frac{π}{2}}^{\frac{π}{2}} 49 cos^{2} (u) d u

Take the constant out:

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

= 49 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 49 \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

= 49 \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

= 49 \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

= 49 \cdot \frac{1}{2} (π + 0)

Simplify

49 \cdot \frac{1}{2} (π + 0) : \frac{49 π}{2}

= \frac{49 π}{2}

\int - 2 x d x

\frac{d}{d x} (\frac{40}{0.01 x ^{2} + 1})

\frac{d}{d x} (e^{n x^{2}})

x^{^{'}} = x^{2} + x

\frac{d}{d x} (lo g_{7} (e^{- x} cos (π x)))

What is the integral from-7 to 7 of (sqrt(49-x^2)) ?
The integral from-7 to 7 of (sqrt(49-x^2)) is (49pi)/2