What is area x=-3,x=3,y=2e^{2x},y=e^{2x}+1 ?

The answer to area x=-3,x=3,y=2e^{2x},y=e^{2x}+1 is (e^{12}-2e^6+1)/(2e^6)

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area x=-3,x=3,y=2e^{2x},y=e^{2x}+1

area

x = - 3, x = 3, y = 2 e^{2 x}, y = e^{2 x} + 1

\frac{e ^{12} - 2 e ^{6} + 1}{2 e ^{6}}

Solution steps

Apply the area formula:

\int_{- 3}^{3} 2 e^{2 x} - (e^{2 x} + 1) d x

Solve

\int_{- 3}^{3} 2 e^{2 x} - (e^{2 x} + 1) d x : \frac{e ^{12} - 2 e ^{6} + 1}{2 e ^{6}}

The area is:

= \frac{e ^{12} - 2 e ^{6} + 1}{2 e ^{6}}

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What is area x=-3,x=3,y=2e^{2x},y=e^{2x}+1 ?
The answer to area x=-3,x=3,y=2e^{2x},y=e^{2x}+1 is (e^{12}-2e^6+1)/(2e^6)