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 $n\neq 0$.

 Solution:

Step 1:
(a) Use a u-substitution with $u=3x^{2}+1.$ This means $du=6x\,dx$, or $dx=du/6$. After substitution we have $\int xe^{3x^{2}+1}\,dx={\frac {1}{6}}\int e^{u}\,du.$
(b) We need to use the power rule to find that $\int _{2}^{5}4x-5\,dx=2x^{2}-5x{\Bigr |}_{2}^{5}$

Step 2:
(a) ${\frac {1}{6}}\int e^{u}\,du={\frac {1}{6}}e^{u}.$
(b) We just need to evaluate at the endpoints to finish the problem: ${\begin{array}{rcl}2x^{2}-5x{\Bigr |}_{2}^{5}&=&2(5^{2})-5(5)-(2(2)^{2}-5(2)\\&=&50-25-(8-10)\\&=&25+2\\&=&27\end{array}}$

Step 3:
(a) Now we need to substitute back into our original variables using our original substitution $u=3x^{2}+1$
to find  ${\frac {1}{6}}e^{u}={\frac {e^{3x^{2}+1}}{6}}.$

Step 4:
Since this integral is an indefinite integral we have to remember to add a constant  $C$ at the end.

Final Answer:
(a) ${\frac {e^{3x^{2}+1}}{6}}+C.$
(b) $27$

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