What is the integral from 0 to 40 of 3000e^{0.04t} ?

The integral from 0 to 40 of 3000e^{0.04t} is 296477.43182…

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integral from 0 to 40 of 3000e^{0.04t}

\int_{0}^{40} 3000 e^{0.04 t} d t

296477.43182 \dots

Solution steps

\int_{0}^{40} 3000 e^{0.04 t} d t

Take the constant out:

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

= 3000 \cdot \int_{0}^{40} e^{0.04 t} d t

Apply u-substitution

: \int_{0}^{1.6} e^{u} \frac{1}{0.04} d u

= 3000 \cdot \int_{0}^{1.6} e^{u} \frac{1}{0.04} d u

Take the constant out:

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

= 3000 \cdot \frac{1}{0.04} \cdot \int_{0}^{1.6} e^{u} d u

Simplify

= 3000 \cdot 25 \cdot \int_{0}^{1.6} e^{u} d u

Use the common integral:

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

= 3000 \cdot 25 [e^{u}]_{0}^{1.6}

Simplify

= 75000 [e^{u}]_{0}^{1.6}

Compute the boundaries:

3.95303 \dots

= 75000 \cdot 3.95303 \dots

Simplify

= 296477.43182 \dots

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What is the integral from 0 to 40 of 3000e^{0.04t} ?
The integral from 0 to 40 of 3000e^{0.04t} is 296477.43182…