Question

evaluate the integral.
int_{0}^{5} e^{3 x} d x
int_{0}^{5} e^{3 x} d x=square
(type an integer or decimal rounded to the nearest thousandth as needed.)

Explanation:

Step1: Use substitution method

Let \( u = 3x \), then \( du=3dx \), \( dx=\frac{1}{3}du \). When \( x = 0 \), \( u = 0 \); when \( x = 5 \), \( u = 15 \).
The integral \(\int_{0}^{5}e^{3x}dx\) becomes \(\frac{1}{3}\int_{0}^{15}e^{u}du\).

Step2: Integrate \(e^{u}\)

We know that \(\int e^{u}du=e^{u}+C\). So \(\frac{1}{3}\int_{0}^{15}e^{u}du=\frac{1}{3}[e^{u}]_{0}^{15}\).

Step3: Evaluate the definite - integral

Using the fundamental theorem of calculus \(F(b)-F(a)\), we have \(\frac{1}{3}(e^{15}-e^{0})\).
Since \(e^{0} = 1\), then \(\frac{1}{3}(e^{15}-1)\).
Calculating \(e^{15}\approx3269017.372\), \(\frac{1}{3}(3269017.372 - 1)=\frac{3269016.372}{3}=1089672.124\).

Answer:

\(1089672.124\)