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\problemgrouptitle{Integration Review}

\begin{problemblock}{Find the antiderivative of the given functions.}

\item $\displaystyle f(x)=\sqrt[5]{x^3}+\frac{4}{\sqrt{x}}+x^3+10$
\answer{$\displaystyle F(x)=\frac{5}{8}x^{8/5}+8x^{1/2}+\frac{1}{4}x^4+10x+C$}

\item $\displaystyle g(x)=\frac{\sqrt{x^3}+5x+1}{2x^3}$
\answer{$\displaystyle G(x)=-x^{-1/2}-\frac{5}{2x}-\frac{1}{4x^2}+C$}

\item $\displaystyle f(\theta)=3\sin\theta+5\cos\theta+\theta^3+1$
\answer{$\displaystyle F(\theta)=-3\cos\theta+5\sin\theta+\frac{1}{4}\theta^4+\theta+C$}

\item $\displaystyle g(\theta)=2\tan\theta\sec\theta-2\cos\theta +\frac{1}{\cos^2\theta}$
\answer{$\displaystyle G(\theta)=2\sec\theta-2\sin\theta+\tan\theta+C$}
\end{problemblock}

\begin{problemblock}{Find function $f$ satisfying the given conditions.}

\item $\displaystyle f''(x)=\sqrt{x}+x^2-6$
\answer{$\displaystyle f(x)=\frac{4}{15}x^{5/2}+\frac{1}{12}x^4-3x^2+Cx+D$}

\item $\displaystyle f''(t)=20\sqrt[3]{x^2}-3x-5$, $f(1)=1$, $f'(1)=-1$
\answer{$\displaystyle f(t)=\frac{9}{2}t^{8/3}-\frac{1}{2}t^3-\frac{5}{2}t^2-\frac{13}{2}t+6$}

\item $\displaystyle f''(\theta )=5\sin\theta -4\cos\theta+10$, $f(0)=-13$, $f'(0)=2$
\answer{$\displaystyle f(\theta)=-5\sin\theta+4\cos\theta+5\theta^2+7\theta-17$}
\end{problemblock}

\begin{problemblock}{Evaluate the given integrals.}

\item $\displaystyle \int x^4\left(x^5+3\right)^8\,dx$
\answer{$\displaystyle \frac{1}{45}\left(x^5+3\right)^9+C$}

\item $\displaystyle \int \sec^2(2\theta)\left[\tan(2\theta)+1\right]^5\,d\theta$
\answer{$\displaystyle \frac{1}{12}\left[\tan(2\theta) +1\right]^6+C$}

\item $\displaystyle \int \frac{\left(\sqrt[5]{t^2}-4\right)^{\frac{1}{3}}}{\sqrt[5]{t^3}}\,dt$
\answer{$\displaystyle \frac{15}{8}\left(t^{2/5}-4\right)^{4/3}+C$}

\item $\displaystyle \int_{-1}^{1}x(x-1)^6\,dx$
\answer{$\displaystyle -\frac{96}{7}$}

\item $\displaystyle \int_{0}^{\pi/4}\cos^4(3x)\sin(3x)\,dx$
\answer{$\displaystyle \frac{1}{15}\left(\frac{\sqrt{2}}{8}+1\right)$}

\item $\displaystyle \int_{0}^{\frac{\pi}{3}}\cos^2(3\theta)\,d\theta$
\answer{$\displaystyle \frac{\pi}{6}$}

\end{problemblock}

\begin{problemblock}{Find the area of the region bounded by the given curves.}

\item $y=x^2+1$ and $y=5$.
\answer{$\displaystyle \frac{32}{3}$}

\item $y=x$, $y=4x$,  and $x+y=3$.
\answer{$\displaystyle \frac{27}{20}$}

\item $x=10-y^2$ and $x=2+y^2$.
\answer{$\displaystyle \frac{64}{3}$}

\end{problemblock}

\begin{problemblock}{Solve the following differential equations and initial value problems.}

\item $\displaystyle y'=\frac{3+x^5}{x^2}$
\answer{$\displaystyle y=-\frac{3}{x}+\frac{1}{4}x^4+C$}
\item $y'\sec x=2$, $y(0)=3$
\answer{$y=2\sin x+3$}

\end{problemblock}
