Question

evaluate the following definite integral to two decimal places.

$$\int_{0}^{8} e^{0.06(8 - t)} d t$$

$$\int_{0}^{8} e^{0.06(8 - t)} d t=\square$$ (round to two decimal places as needed.)

Explanation:

Step1: Use substitution

Let \(u = 0.06(8 - t)=0.48-0.06t\), then \(du=- 0.06dt\), and \(dt=-\frac{du}{0.06}\). When \(t = 0\), \(u = 0.48\); when \(t = 8\), \(u=0\).
The integral \(\int_{0}^{8}e^{0.06(8 - t)}dt\) becomes \(-\frac{1}{0.06}\int_{0.48}^{0}e^{u}du\).

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

We know that \(\int e^{u}du=e^{u}+C\). So \(-\frac{1}{0.06}\int_{0.48}^{0}e^{u}du=\frac{1}{0.06}\int_{0}^{0.48}e^{u}du\).
Using the fundamental theorem of calculus \(F(b)-F(a)\) for \(\int_{a}^{b}f(x)dx\) where \(F(x)\) is an antiderivative of \(f(x)\), we have \(\frac{1}{0.06}[e^{u}]_{0}^{0.48}\).

Step3: Evaluate the definite - integral

\(\frac{1}{0.06}(e^{0.48}-e^{0})\). Since \(e^{0}=1\), and \(e^{0.48}\approx1.61616\) (using \(e^{x}\approx1 + x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots\) or a calculator), then \(\frac{1}{0.06}(1.61616 - 1)=\frac{0.61616}{0.06}\approx10.27\).

Answer:

\(10.27\)