Question

1. 10 pts, 5 each compute the following integrals using the fundamental theorem of calculus:

• (int_{-1}^{2}(10x^{4}-3x^{2}+1)dx)
• (int_{8/27}^{1}\frac{8t^{1/3}-10t^{4/3}}{t^{2}}dt)

Explanation:

Step1: Integrate the first integral

Integrate \(10x^{4}-3x^{2}+1\) term - by - term.
Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)):
\(\int(10x^{4}-3x^{2}+1)dx=10\times\frac{x^{5}}{5}-3\times\frac{x^{3}}{3}+x+C = 2x^{5}-x^{3}+x+C\)
Then apply the Fundamental Theorem of Calculus \(\int_{a}^{b}F^\prime(x)dx=F(b)-F(a)\)
\(F(x)=2x^{5}-x^{3}+x\), \(a=-1\), \(b = 2\)
\(F(2)-F(-1)=(2\times2^{5}-2^{3}+2)-(2\times(-1)^{5}-(-1)^{3}+(-1))\)
\(=(2\times32 - 8+2)-(-2 + 1-1)\)
\(=(64-8 + 2)-(-2)\)
\(=58 + 2=60\)

Step2: Simplify and integrate the second integral

First, simplify \(\frac{8t^{1/3}-10t^{4/3}}{t^{2}}=8t^{\frac{1}{3}-2}-10t^{\frac{4}{3}-2}=8t^{-\frac{5}{3}}-10t^{-\frac{2}{3}}\)
Integrate term - by - term using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\))
\(\int(8t^{-\frac{5}{3}}-10t^{-\frac{2}{3}})dt=8\times\frac{t^{-\frac{5}{3}+1}}{-\frac{5}{3}+1}-10\times\frac{t^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+C\)
\(=8\times\frac{t^{-\frac{2}{3}}}{-\frac{2}{3}}-10\times\frac{t^{\frac{1}{3}}}{\frac{1}{3}}+C=-12t^{-\frac{2}{3}}-30t^{\frac{1}{3}}+C\)
Apply the Fundamental Theorem of Calculus with \(a=\frac{8}{27}\), \(b = 1\)
\(F(t)=-12t^{-\frac{2}{3}}-30t^{\frac{1}{3}}\)
\(F(1)-F(\frac{8}{27})=(-12\times1^{-\frac{2}{3}}-30\times1^{\frac{1}{3}})-(-12\times(\frac{8}{27})^{-\frac{2}{3}}-30\times(\frac{8}{27})^{\frac{1}{3}})\)
Since \((\frac{8}{27})^{\frac{1}{3}}=\frac{2}{3}\) and \((\frac{8}{27})^{-\frac{2}{3}}=(\frac{27}{8})^{\frac{2}{3}}=\frac{9}{4}\)
\(=(-12-30)-(-12\times\frac{9}{4}-30\times\frac{2}{3})\)
\(=-42-(-27 - 20)\)
\(=-42 + 47=5\)

Answer:

The value of \(\int_{-1}^{2}(10x^{4}-3x^{2}+1)dx\) is \(60\) and the value of \(\int_{\frac{8}{27}}^{1}\frac{8t^{1/3}-10t^{4/3}}{t^{2}}dt\) is \(5\)