What is the integral from 0 to 2 of 10e^{2x} ?

The integral from 0 to 2 of 10e^{2x} is 5(e^4-1)

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integral from 0 to 2 of 10e^{2x}

\int_{0}^{2} 10 e^{2 x} d x

5 (e^{4} - 1)

Decimal Notation

267.99075 \dots

Solution steps

\int_{0}^{2} 10 e^{2 x} d x

Take the constant out:

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

= 10 \cdot \int_{0}^{2} e^{2 x} d x

Apply u-substitution

: \int_{0}^{4} e^{u} \frac{1}{2} d u

= 10 \cdot \int_{0}^{4} e^{u} \frac{1}{2} d u

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \int_{0}^{4} e^{u} d u

Use the common integral:

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

= 10 \cdot \frac{1}{2} [e^{u}]_{0}^{4}

Simplify

10 \cdot \frac{1}{2} [e^{u}]_{0}^{4} : 5 [e^{u}]_{0}^{4}

= 5 [e^{u}]_{0}^{4}

Compute the boundaries:

e^{4} - 1

= 5 (e^{4} - 1)

\int 3 x^{2} arctan (x^{3}) d x

x \to 3 + lim (9 - x^{2})

\frac{\partial}{\partial x} (e^{3 x} sin (2 y))

x^{2} - 3 x + 7

f (x) = x^{2} + 4, g (x) = - x^{2} - 1, [- 2, 3]

What is the integral from 0 to 2 of 10e^{2x} ?
The integral from 0 to 2 of 10e^{2x} is 5(e^4-1)