What is the integral from 0 to 1 of (e^{(2x)/3}+1) ?

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

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integral from 0 to 1 of (e^{(2x)/3}+1)

\int_{0}^{1} (e^{\frac{2 x}{3}} + 1) d x

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

Decimal

2.42160 \dots

Solution steps

\int_{0}^{1} e^{\frac{2 x}{3}} + 1 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^{\frac{2 x}{3}} d x + \int_{0}^{1} 1 d x

\int_{0}^{1} e^{\frac{2 x}{3}} d x = \frac{3 ( e ^{\frac{2}{3}} - 1 )}{2}

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

= \frac{3 ( e ^{\frac{2}{3}} - 1 )}{2} + 1

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