What is the integral from 0 to 1 of (e^{2x}-e^x) ?

The integral from 0 to 1 of (e^{2x}-e^x) is (e^2-1)/2-e+1

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integral from 0 to 1 of (e^{2x}-e^x)

\int_{0}^{1} (e^{2 x} - e^{x}) d x

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

Decimal Notation

1.47624 \dots

Solution steps

\int_{0}^{1} e^{2 x} - e^{x} 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_{0}^{1} e^{2 x} d x - \int_{0}^{1} e^{x} d x

\int_{0}^{1} e^{2 x} d x = \frac{e ^{2} - 1}{2}

\int_{0}^{1} e^{x} d x = e - 1

= \frac{e ^{2} - 1}{2} - (e - 1)

Simplify

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

y^{^{''}} + 64 y^{^{'}} = 0

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

y^{^{'}} + x^{2} sec (y) = 0

- 5

\int 50 + 25 e^{10 t} + 25 e^{- 10 t} d t

What is the integral from 0 to 1 of (e^{2x}-e^x) ?
The integral from 0 to 1 of (e^{2x}-e^x) is (e^2-1)/2-e+1