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 6e^4(\sqrt[3]{e}-1)+5

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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

Decimal Notation

134.59823 \dots

Solution steps

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

e^{\frac{x}{3} + 4} = e^{4} e^{\frac{x}{3}}

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

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

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

\int_{1}^{e^{3}} \frac{ln ( x )}{x} d x

\int_{0}^{1} x^{2} (5 x^{3} - 1)^{4} d x

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

\int_{0}^{π} x sin (x^{2}) d x

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 6e^4(\sqrt[3]{e}-1)+5