Question

find the area under the graph of the function for the given interval. (round your answer to two decimal places.)
y = 8x + 0.2e^x, 1 ≤ x ≤ 4

Explanation:

Step1: Recall the area - under - curve formula

The area \(A\) under the curve \(y = f(x)\) from \(x=a\) to \(x = b\) is given by \(A=\int_{a}^{b}f(x)dx\). Here, \(f(x)=8x + 0.2e^{x}\), \(a = 1\), and \(b = 4\), so \(A=\int_{1}^{4}(8x + 0.2e^{x})dx\).

Step2: Use the integral sum - rule

\(\int_{1}^{4}(8x + 0.2e^{x})dx=\int_{1}^{4}8xdx+\int_{1}^{4}0.2e^{x}dx\).

Step3: Integrate \(8x\)

\(\int_{1}^{4}8xdx=8\int_{1}^{4}xdx\). Using the power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have \(8\times[\frac{x^{2}}{2}]_{1}^{4}=4[x^{2}]_{1}^{4}=4(4^{2}-1^{2})=4(16 - 1)=60\).

Step4: Integrate \(0.2e^{x}\)

\(\int_{1}^{4}0.2e^{x}dx=0.2\int_{1}^{4}e^{x}dx\). Since \(\int e^{x}dx=e^{x}+C\), then \(0.2[e^{x}]_{1}^{4}=0.2(e^{4}-e^{1})\).
\(e\approx2.7183\), \(e^{4}\approx54.5982\), so \(0.2(e^{4}-e^{1})=0.2(54.5982 - 2.7183)=0.2\times51.8799 = 10.37598\).

Step5: Find the total area

\(A = 60+10.37598=70.37598\approx70.38\).

Answer:

\(70.38\)