What is the integral from 0 to 2ln(7) of (5e^{x/2}) ?

The integral from 0 to 2ln(7) of (5e^{x/2}) is 60

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integral from 0 to 2ln(7) of (5e^{x/2})

\int_{0}^{2 l n (7)} (5 e^{\frac{x}{2}}) d x

60

Solution steps

\int_{0}^{2 l n (7)} 5 e^{\frac{x}{2}} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{2 l n (7)} e^{\frac{x}{2}} d x

Apply u-substitution

: \int_{0}^{l n (7)} e^{u} \cdot 2 d u

= 5 \cdot \int_{0}^{l n (7)} e^{u} \cdot 2 d u

Take the constant out:

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

= 5 \cdot 2 \cdot \int_{0}^{l n (7)} e^{u} d u

Use the common integral:

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

= 5 \cdot 2 [e^{u}]_{0}^{l n (7)}

Simplify

= 10 [e^{u}]_{0}^{l n (7)}

Compute the boundaries:

6

= 10 \cdot 6

Simplify

= 60

6 x^{2} - 5

\frac{\partial}{\partial x} (sin (5 x + 5 y))

n = 1 \sum \infty sin^{2} (n)

f (x) = x, (81, 9)

3 e^{- 5 t} + 6 e^{- 3 t}

What is the integral from 0 to 2ln(7) of (5e^{x/2}) ?
The integral from 0 to 2ln(7) of (5e^{x/2}) is 60