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\problemgrouptitle{Indefinite Integrals}

\item Explain why we use the indefinite integral symbol, $\int\! f(x)\,dx$, to represent the general form of the antiderivative of the function $f(x)$\,.
\answer{The Fundamental Theorem of Calculus shows that the definite integral $\int_a^b\!f(x)\,dx$ is, assuming the conditions of the theorem are met, intimately connected to the antiderivative of $f$ by the relation $\int_a^b\!f(x)\,dx=F(b)-F(a)$ where $F$ is an antiderivative of $f$.  Thus in many cases finding a definite integral is a two-step process where first one finds an antiderivative of $f$ and then secondly takes the difference of that function evaluated at the limits of integration.  It is natural, therefore, to generally write the answer to the first step, namely the antiderivative of $f$, symbolically as $\int f(x)\,dx$.  (The notation is further convenient because it embeds the function we are antidifferentiating directly in the symbol in the same way we write $\frac{df}{dx}$ abstractly for the derivative of $f$.)}  

\begin{problemblock}[2]{Evaluate the following indefinite integrals.}

\item $\displaystyle\int \left(x^3-3x^4-6\right)\d{x}$
\answer{$\displaystyle\frac{x^4}{4}-\frac{3x^5}{5}-6x+C$}

\item $\displaystyle\int\frac{2+x}{\sqrt{x}}\d{x}$
\answer{$\displaystyle 4x^{\frac{1}{2}}+\frac{2}{3}x^{\frac{3}{2}}+C$}

\item $\displaystyle\int\csc \theta\cot \theta\d{\theta}$
\answer{$-\csc \theta+C$}

\item $\displaystyle\int \left(\tan^2x+1\right)\d{x}$
\answer{$\displaystyle\int\left(\tan^2x+1\right)\d{x}=\int\sec^2x\d{x} =\tan x+C$}

\end{problemblock}

\item Find the general form of the function $y=f(x)$ such that the equation $y'=x^2+9$ is satisfied.
\answer{The equation states that the derivative of $y$ is $x^2+9$ so $y$ must be the general form of the antiderivative of $x^2+9$ which is just its indefinite integral:

$\displaystyle y=\int\!\left(x^2+9\right)\,dx=\frac{1}{3}x^3+9x+C$

The general solution for any differential equation of the form $y'=f(x)$ is similarly \mbox{$y=\int f(x)\,dx$\,.}}
