What is area f(x)=-x^2+2,g(x)=-x-4 ?

The answer to area f(x)=-x^2+2,g(x)=-x-4 is 125/6

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area f(x)=-x^2+2,g(x)=-x-4

area

f (x) = - x^{2} + 2, g (x) = - x - 4

\frac{125}{6}

Solution steps

Apply the area formula:

\int_{- 2}^{3} - x^{2} + 2 - (- x - 4) d x

Solve

\int_{- 2}^{3} - x^{2} + 2 - (- x - 4) d x : \frac{125}{6}

The area is:

= \frac{125}{6}

\int \frac{- 2}{x + 1} d x

\frac{d}{d x} (\frac{1}{cos ^{2} ( x )})

n = 1 \sum \infty n^{- \frac{3}{2}}

\int v^{- \frac{4}{5}} d v

e^{- i x}

What is area f(x)=-x^2+2,g(x)=-x-4 ?
The answer to area f(x)=-x^2+2,g(x)=-x-4 is 125/6