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

The integral from 0 to 1 of (2+e^x)^2 is 4e+(e^2-1)/2

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

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

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

Solution steps

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

Expand

(2 + e^{x})^{2} : 4 + 4 e^{x} + e^{2 x}

= \int_{0}^{1} 4 + 4 e^{x} + e^{2 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} 4 d x + \int_{0}^{1} 4 e^{x} d x + \int_{0}^{1} e^{2 x} d x

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

\int_{0}^{1} 4 e^{x} d x = 4 (e - 1)

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

= 4 + 4 (e - 1) + \frac{e ^{2} - 1}{2}

Simplify

4 + 4 (e - 1) + \frac{e ^{2} - 1}{2} : 4 e + \frac{e ^{2} - 1}{2}

= 4 e + \frac{e ^{2} - 1}{2}

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

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

\int_{1}^{3} (5 x - x^{2}) d x

\int_{4}^{5} x^{2} d x

\int_{0}^{4} (4 - x) (x) d x

What is the integral from 0 to 1 of (2+e^x)^2 ?
The integral from 0 to 1 of (2+e^x)^2 is 4e+(e^2-1)/2