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}
