What is the integral from-7 to 7 of (1+sqrt(49-x^2)) ?

The integral from-7 to 7 of (1+sqrt(49-x^2)) is 14+(49pi)/2

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integral from-7 to 7 of (1+sqrt(49-x^2))

\int_{- 7}^{7} (1 + 49 - x^{2}) d x

14 + \frac{49 π}{2}

Decimal

90.96902 \dots

Solution steps

\int_{- 7}^{7} 1 + 49 - x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 7}^{7} 1 d x + \int_{- 7}^{7} 49 - x^{2} d x

\int_{- 7}^{7} 1 d x = 14

\int_{- 7}^{7} 49 - x^{2} d x = \frac{49 π}{2}

= 14 + \frac{49 π}{2}

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

f (x) = \frac{1}{( x - 5 )}, at x = - 3

\frac{2 s + 1}{s ^{2} - 2 s + 2}

\int_{2}^{3} 3 x^{2} d x

2 x + 3

What is the integral from-7 to 7 of (1+sqrt(49-x^2)) ?
The integral from-7 to 7 of (1+sqrt(49-x^2)) is 14+(49pi)/2