What is area y= 1/16 (e^{8x}+e^{-8x}),-4<= x<= 4 ?

The answer to area y= 1/16 (e^{8x}+e^{-8x}),-4<= x<= 4 is (e^{64}-1)/(64e^{32)}

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area y= 1/16 (e^{8x}+e^{-8x}),-4<= x<= 4

area

y = \frac{1}{16} (e^{8 x} + e^{- 8 x}), - 4 \leq x \leq 4

\frac{e ^{64} - 1}{64 e ^{32}}

Decimal Notation

1233796252854.3838

Solution steps

Apply the area formula:

\int_{- 4}^{4} \frac{1}{16} e^{8 x} + e^{- 8 x} d x

Solve

\int_{- 4}^{4} \frac{1}{16} e^{8 x} + e^{- 8 x} d x : \frac{e ^{64} - 1}{64 e ^{32}}

The area is:

= \frac{e ^{64} - 1}{64 e ^{32}}

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

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\frac{d}{d x} ((4 x^{3} + 5 x) (x - 3) (x + 5))

\frac{d}{d x} (5 arcsin (3) x^{3})

What is area y= 1/16 (e^{8x}+e^{-8x}),-4<= x<= 4 ?
The answer to area y= 1/16 (e^{8x}+e^{-8x}),-4<= x<= 4 is (e^{64}-1)/(64e^{32)}