022 Exam 2 Sample B, Problem 7

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Find the antiderivatives:

(a) $\int xe^{3x^{2}+1}\,dx.$ (b) $\int _{2}^{5}4x-5\,dx.$ Foundations:
This problem requires Integration by substitution (u - sub): If $u=g(x)$ is a differentiable functions whose range is in the domain of $f$ , then
$\int g'(x)f(g(x))dx\,=\,\int f(u)du.$ We also need our power rule for integration:
$\int x^{n}dx\,=\,{\frac {x^{n+1}}{n+1}}+C,$ for $n\neq -1$ ,
as well as the convenient antiderivative:
$\int e^{x}\,dx\,=\,e^{x}+C.$ Solution:

(a) Step 1:
(a) Use a u-substitution with $u=3x^{2}+1.$ This means $du=6x\,dx$ , or $dx=du/6$ . Substituting, we have
$\int xe^{3x^{2}+1}\,dx\,=\,\int xe^{u}\cdot {\frac {du}{6}}\,=\,{\frac {1}{6}}\int e^{u}\,du\,=\,{\frac {1}{6}}u.$ (a) Step 2:
Now, we need to substitute back into our original variable using our original substitution $u=3x^{2}+1$ to find  ${\frac {1}{6}}e^{u}={\frac {e^{3x^{2}+1}}{6}}.$ (a) Step 3:
Since this integral is an indefinite integral, we have to remember to add a constant  $C$ at the end.
(b):
Unlike part (a), this requires no substitution. We can integrate term-by-term to find
$\int _{2}^{5}4x-5\,dx=2x^{2}-5x{\Bigr |}_{x\,=\,2}^{5}.$ Then, we evaluate:
${\begin{array}{rcl}2x^{2}-5x{\Bigr |}_{x\,=\,2}^{5}&=&2(5^{2})-5(5)-(2(2)^{2}-5(2))\\&=&50-25-(8-10)\\&=&25+2\\&=&27.\end{array}}$ $\int xe^{3x^{2}+1}\,dx\,=\,{\frac {e^{3x^{2}+1}}{6}}+C.$ $\int _{2}^{5}4x-5\,dx\,=\,27.$ 