What is the integral from-7 to 0 of (4+sqrt(49-x^2)) ?

The integral from-7 to 0 of (4+sqrt(49-x^2)) is 28+(49pi)/4

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integral from-7 to 0 of (4+sqrt(49-x^2))

\int_{- 7}^{0} (4 + 49 - x^{2}) d x

28 + \frac{49 π}{4}

Decimal Notation

66.48451 \dots

Solution steps

\int_{- 7}^{0} 4 + 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}^{0} 4 d x + \int_{- 7}^{0} 49 - x^{2} d x

\int_{- 7}^{0} 4 d x = 28

\int_{- 7}^{0} 49 - x^{2} d x = \frac{49 π}{4}

= 28 + \frac{49 π}{4}

\int \frac{2 x ^{2} + 5 x + 2}{( x ^{2} + 1 ) ^{2}} d x

e^{y^{2}} d x + x^{2} y d y = 0

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

\int \frac{25 - 16 x ^{2}}{x} d x

What is the integral from-7 to 0 of (4+sqrt(49-x^2)) ?
The integral from-7 to 0 of (4+sqrt(49-x^2)) is 28+(49pi)/4