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

The integral from 0 to 1 of (2x+1)/(e^x) is -5/e+3

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integral from 0 to 1 of (2x+1)/(e^x)

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

- \frac{5}{e} + 3

Decimal

1.16060 \dots

Solution steps

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

Expand

\frac{2 x + 1}{e ^{x}} : \frac{2 x}{e ^{x}} + \frac{1}{e ^{x}}

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

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

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

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

Simplify

2 (- \frac{2}{e} + 1) - \frac{1}{e} + 1 : - \frac{5}{e} + 3

= - \frac{5}{e} + 3

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What is the integral from 0 to 1 of (2x+1)/(e^x) ?
The integral from 0 to 1 of (2x+1)/(e^x) is -5/e+3