What is area y=20-x^2\land y=2x+5,-2<= x<= 8 ?

The answer to area y=20-x^2\land y=2x+5,-2<= x<= 8 is 200

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area y=20-x^2\land y=2x+5,-2<= x<= 8

area

y = 20 - x^{2} and y = 2 x + 5, - 2 \leq x \leq 8

200

Solution steps

Apply the area formula:

\int_{- 2}^{8} 20 - x^{2} - (2 x + 5) d x

Solve

\int_{- 2}^{8} 20 - x^{2} - (2 x + 5) d x : 200

The area is:

= 200

\int \frac{22}{( 121 x ^{2} + 1 ) ^{2}} d x

\frac{\partial}{\partial x} (x yz - cos (x + y + z))

\int - g e^{- c t} \frac{1}{c} d t

\frac{\partial}{\partial x} (4 x^{2} + 2 x y - 8 y^{2})

\frac{\partial}{\partial x} (- \frac{y}{x ^{2} + y ^{2}})

What is area y=20-x^2\land y=2x+5,-2<= x<= 8 ?
The answer to area y=20-x^2\land y=2x+5,-2<= x<= 8 is 200