# 022 Exam 2 Sample B, Problem 7

Find the antiderivatives:

(a) ${\displaystyle \int xe^{3x^{2}+1}\,dx.}$

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

Solution:

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