What is the integral from-5 to 0 of (4+sqrt(25-x^2)) ?

The integral from-5 to 0 of (4+sqrt(25-x^2)) is 20+(25pi)/4

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integral from-5 to 0 of (4+sqrt(25-x^2))

\int_{- 5}^{0} (4 + 25 - x^{2}) d x

20 + \frac{25 π}{4}

Decimal Notation

39.63495 \dots

Solution steps

\int_{- 5}^{0} 4 + 25 - 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_{- 5}^{0} 4 d x + \int_{- 5}^{0} 25 - x^{2} d x

\int_{- 5}^{0} 4 d x = 20

\int_{- 5}^{0} 25 - x^{2} d x = \frac{25 π}{4}

= 20 + \frac{25 π}{4}

x \to 8 lim (\frac{x + 9}{x - 8})

y^{^{''}} + 256 y = 7 cos (5 t)

\frac{10}{s ^{2} + 25} + \frac{4}{s - 3}

\int \frac{x ^{2} - 25}{x} d x

\int 1064 - 0.005 x d x

What is the integral from-5 to 0 of (4+sqrt(25-x^2)) ?
The integral from-5 to 0 of (4+sqrt(25-x^2)) is 20+(25pi)/4