022 Exam 2 Sample B, Problem 7

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 $n\neq 0$,
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:
${\begin{array}{rcl}2x^{2}-5x{\Bigr |}_{x\,=\,2}^{5}&=&2(5^{2})-5(5)-(2(2)^{2}-5(2))\\&=&50-25-(8-10)\\&=&25+2\\&=&27.\end{array}}$

Final Answer:
(a) $\int xe^{3x^{2}+1}\,dx\,=\,{\frac {e^{3x^{2}+1}}{6}}+C.$
(b)
$\int _{2}^{5}4x-5\,dx\,=\,27.$

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