Question

let the region r be the area enclosed by the function $f(x) = e^x$ and $g(x) = 5x + 1$. if the region r is the base of a solid such that each cross section perpendicular to the $x$-axis is a rectangle whose height is half the length of its base in the region r, find the volume of the solid. you may use a calculator and round to the nearest thousandth.

Explanation:

Step1: Find intersection points

Set \(e^{x}=5x + 1\). Using a calculator, the intersection points are \(x = 0\) and \(x\approx2.25\).

Step2: Determine the base and height of the cross - section

The base of the rectangle \(b(x)=g(x)-f(x)=(5x + 1)-e^{x}\). The height \(h(x)=\frac{1}{2}b(x)=\frac{1}{2}[(5x + 1)-e^{x}]\).

Step3: Use the volume formula for cross - sectional area

The volume formula \(V=\int_{a}^{b}A(x)dx\), where \(A(x)\) is the area of the cross - section. Since \(A(x)=b(x)\times h(x)\), and \(h(x)=\frac{1}{2}b(x)\), then \(A(x)=\frac{1}{2}[b(x)]^{2}\). So \(A(x)=\frac{1}{2}[(5x + 1)-e^{x}]^{2}\).

Step4: Calculate the integral

\(V=\frac{1}{2}\int_{0}^{2.25}[(5x + 1)-e^{x}]^{2}dx\)
Expand \([(5x + 1)-e^{x}]^{2}=(5x + 1)^{2}-2(5x + 1)e^{x}+e^{2x}=25x^{2}+10x + 1-10xe^{x}-2e^{x}+e^{2x}\)
\(V=\frac{1}{2}\int_{0}^{2.25}(25x^{2}+10x + 1-10xe^{x}-2e^{x}+e^{2x})dx\)
We know that:

• \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n

eq - 1)\), so \(\int_{0}^{2.25}25x^{2}dx=25\times\frac{x^{3}}{3}\big|_{0}^{2.25}\)

• \(\int_{0}^{2.25}10xdx = 10\times\frac{x^{2}}{2}\big|_{0}^{2.25}\)
• \(\int_{0}^{2.25}1dx=x\big|_{0}^{2.25}\)
• Use integration by parts \(\int u dv=uv-\int v du\) for \(\int xe^{x}dx\) (let \(u = x\), \(dv=e^{x}dx\), then \(du = dx\), \(v = e^{x}\), so \(\int xe^{x}dx=xe^{x}-e^{x}+C\)), \(\int_{0}^{2.25}- 10xe^{x}dx=-10[xe^{x}-e^{x}]\big|_{0}^{2.25}\)
• \(\int_{0}^{2.25}-2e^{x}dx=-2e^{x}\big|_{0}^{2.25}\)
• \(\int_{0}^{2.25}e^{2x}dx=\frac{1}{2}e^{2x}\big|_{0}^{2.25}\)

After calculating each integral:
\(\int_{0}^{2.25}25x^{2}dx=25\times\frac{(2.25)^{3}}{3}\approx25\times\frac{11.390625}{3}\approx94.922\)
\(\int_{0}^{2.25}10xdx=10\times\frac{(2.25)^{2}}{2}=10\times\frac{5.0625}{2} = 25.3125\)
\(\int_{0}^{2.25}1dx=2.25\)
\(\int_{0}^{2.25}-10xe^{x}dx=-10[(2.25e^{2.25}-e^{2.25})-(0 - 1)]\)
\(e^{2.25}\approx9.4877\), \(-10[(2.25\times9.4877-9.4877)+1]=-10[(1.25\times9.4877)+1]\approx-10(11.8596 + 1)\approx-128.596\)
\(\int_{0}^{2.25}-2e^{x}dx=-2(e^{2.25}-1)\approx-2(9.4877 - 1)\approx-16.975\)
\(\int_{0}^{2.25}e^{2x}dx=\frac{1}{2}(e^{4.5}-1)\), \(e^{4.5}\approx90.017\), \(\frac{1}{2}(90.017 - 1)\approx44.508\)

\(\int_{0}^{2.25}(25x^{2}+10x + 1-10xe^{x}-2e^{x}+e^{2x})dx\approx94.922+25.3125 + 2.25-128.596-16.975 + 44.508\approx21.4215\)

\(V=\frac{1}{2}\times21.4215 = 10.711\)

Answer:

\(10.711\)