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

The integral from 0 to 1 of 2e^{x/3+4} is 6e^4(\sqrt[3]{e}-1)

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

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

Decimal

129.59823 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

= 2 e^{4} \cdot \int_{0}^{\frac{1}{3}} e^{u} \cdot 3 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 e^{4} \cdot 3 \cdot \int_{0}^{\frac{1}{3}} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= 2 e^{4} \cdot 3 [e^{u}]_{0}^{\frac{1}{3}}

Simplify

= 6 e^{4} [e^{u}]_{0}^{\frac{1}{3}}

Compute the boundaries:

What is the integral from 0 to 1 of 2e^{x/3+4} ?
The integral from 0 to 1 of 2e^{x/3+4} is 6e^4(\sqrt[3]{e}-1)