What is the integral from 0 to 1 of (2e^{x/(3+4)}+5) ?

The integral from 0 to 1 of (2e^{x/(3+4)}+5) is 14\sqrt[7]{e}-9

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integral from 0 to 1 of (2e^{x/(3+4)}+5)

\int_{0}^{1} (2 e^{\frac{x}{3 + 4}} + 5) d x

14 7 e - 9

Decimal

7.14990 \dots

Solution steps

\int_{0}^{1} 2 e^{\frac{x}{3 + 4}} + 5 d x

Simplify

= \int_{0}^{1} 2 e^{\frac{x}{7}} + 5 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} 2 e^{\frac{x}{7}} d x + \int_{0}^{1} 5 d x

\int_{0}^{1} 2 e^{\frac{x}{7}} d x = 14 (7 e - 1)

\int_{0}^{1} 5 d x = 5

= 14 (7 e - 1) + 5

Simplify

14 (7 e - 1) + 5 : 14 7 e - 9

= 14 7 e - 9

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What is the integral from 0 to 1 of (2e^{x/(3+4)}+5) ?
The integral from 0 to 1 of (2e^{x/(3+4)}+5) is 14\sqrt[7]{e}-9