# Find the Definite Integral of an Exponential Function

Find the definite integral of the exponential function **e ^{√x}**, Where e is the exponential constant

\begin{aligned} \int_{0}^{4}e^{\sqrt x} \ dx = ?& \\ \end{aligned}

## Solution to the definite integral of the exponential function

Let

\begin{aligned} I=\int e^{\sqrt x} \ dx\\ \end{aligned}

Let u = √x then, du = dx/(2√x)

⇒ du = 2√x dx

⇒ dx = 2u du

When x = 0 then, u = √x = √0 = 0

When x = 4 then, u = √x = √4 = 2

Now

\begin{aligned} I=\int_{0}^{4}e^{\sqrt x} \ dx &= \int_{0}^{2}(e^u \times 2u) \ du\\ \end{aligned}

Apply Integrate by parts, then

\begin{aligned} I&= 2\int ue^u \ du \\ \\ &= 2u\int e^u \ du-2\int \bigg(\frac{du}{du} \int e^u \ dx\bigg) \ du \\ \\ &= 2u\int e^u \ du-2\int e^u\ du \\ \\ &= 2u e^u - 2e^u \end{aligned}

Apply limits

\begin{aligned} 2\int_{0}^{2} ue^u \ du&= \bigg[2u e^u - 2e^u \bigg]_{0}^{2} \\ \\ &= 2\times 2 e^2 - 2e^2 -(2\times0 \times e^0 - 2e^0) \\ \\ &= 4 e^2 - 2e^2 + 2 \\ \\ &= 2 e^2 + 2 \\ \\ \end{aligned}

That is

\begin{aligned} \int_{0}^{4}e^{\sqrt x} \ dx = 2 e^2 + 2 \\ \\ \end{aligned}