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

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

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

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

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

Decimal

1.08616 \dots

Solution steps

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

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

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

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

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

= e - 1 - (- \frac{1}{e} + 1)

Simplify

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

\int_{- 4}^{3} ∣ - 6 x ∣ d x

\int_{1}^{4} \frac{1}{1 + 2 x} d x

\int_{1}^{\infty} \frac{7}{1 + x ^{2}} d x

\int_{1 - x}^{x} x^{2} y d y

\int_{0}^{2} 1 4^{2 x} d x

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