Evaluating Definite Integrals (College Board AP® Calculus AB): Revision Note

Evaluate the definite integral ${\int }_{-\frac{\pi }{3}}^{\frac{\pi }{4}}\cos xdx$.

Answer:

$\int \cos xdx=\sin x+C$, so $F\left(x\right)=\sin x$ is the antiderivative we can use to evaluate the definite integral

Use ${\int }_a^{b}f\left(x\right)dx={\left[F\left(x\right)\right]}_a^b=F\left(b\right)-F\left(a\right)$

$\begin{array}{rcl}{\int }_{-\frac{\pi }{3}}^{\frac{\pi }{4}}\cos xdx& =& {\left[\sin x\right]}_{-\frac{\pi }{3}}^{\frac{\pi }{4}}\\ & =& \sin \frac{\pi }{4}-\sin \left(-\frac{\pi }{3}\right)\\ & =& \frac{\sqrt{2}}{2}-\left(-\frac{\sqrt{3}}{2}\right)\end{array}$

${\int }_{-\frac{\pi }{3}}^{\frac{\pi }{4}}\cos xdx=\frac{\sqrt{2}+\sqrt{3}}{2}$

Consider the function $f$ defined by $f\left(x\right)={\int }_0^x{t}^2\left(2-t\right)dt$.

(a) Calculate $f\left(2\right)$ and $f\left(3\right)$ and confirm that $f\left(3\right)<f\left(2\right)$.

(b) By considering the properties of the expression being integrated, explain why you would expect $f\left(3\right)<f\left(2\right)$ to be true.

Answer:

(a)

Start by expanding the brackets and finding the indefinite integral

$\begin{array}{rcl}\int t^2\left(2-t\right)dt& =& \int \left(2t^2-t^3\right)dt\\ & =& \frac{2}{3}{t}^3-\frac{1}{4}{t}^4+C\end{array}$

That gives the antiderivative that can be used for calculating both $f\left(2\right)$ and $f\left(3\right)$

$\begin{array}{rcl}f\left(2\right)& =& {\int }_0^2{t}^2\left(2-t\right)dt\\ & =& {\left[\frac{2}{3}{t}^3-\frac{1}{4}{t}^4\right]}_0^2\\ & =& \left(\frac{2}{3}{\left(2\right)}^3-\frac{1}{4}{\left(2\right)}^4\right)-\left(\frac{2}{3}{\left(0\right)}^3-\frac{1}{4}{\left(0\right)}^4\right)\\ & =& \frac{4}{3}-0\\ & =& \frac{4}{3}\end{array}$

$\begin{array}{rcl}f\left(3\right)& =& {\int }_0^3{t}^2\left(2-t\right)dt\\ & =& {\left[\frac{2}{3}{t}^3-\frac{1}{4}{t}^4\right]}_0^3\\ & =& \left(\frac{2}{3}{\left(3\right)}^3-\frac{1}{4}{\left(3\right)}^4\right)-\left(\frac{2}{3}{\left(0\right)}^3-\frac{1}{4}{\left(0\right)}^4\right)\\ & =& -\frac{9}{4}-0\\ & =& -\frac{9}{4}\end{array}$

$f\left(2\right)=\frac{4}{3}$ and $f\left(3\right)=-\frac{9}{4}$, so $f\left(3\right)<f\left(2\right)$

(b)

Consider the sign of $t^2\left(2-t\right)$ between 0 and 3, and recall that negative values of a function being integrated contribute negative quantities to a definite integral

Note that between 0 and 2, $t^2$ and $2-t$ are both positive; while for $t>2$, $t^2$ is positive and $\left(2-t\right)$ is negative

Between 0 and 2, $t^2\left(2-t\right)>0$, so the value of the definite integral between those values must be positive

For $t>2$, $t^2\left(2-t\right)<0$, so the part of the definite integral from 2 to 3 will be negative, and will subtract from the value found between 0 and 2

Consider the function $f$ defined by $f\left(x\right)=\{\begin{array}{ll}2x+1& x\le 1\\ \frac{1}{x}& x>1\end{array}$.

Evaluate the definite integral ${\int }_{-\frac{1}{2}}^{3}f\left(x\right)dx$.

Answer:

This function is not continuous at $x=1$, because $\underset{x\to 1^{-}}{\lim }f\left(x\right)=3$ and $\underset{x\to 1^{+}}{\lim }f\left(x\right)=1$

However, that is a jump discontinuity, so we can just integrate the function separately for each of the pieces

$\begin{array}{rcl}{\int }_{-\frac{1}{2}}^{3}f\left(x\right)dx& =& {\int }_{-\frac{1}{2}}^{1}f\left(x\right)dx+{\int }_1^{3}f\left(x\right)dx\\ & =& {\int }_{-\frac{1}{2}}^1\left(2x+1\right)dx+{\int }_1^3\frac{1}{x}dx\\ & =& {\left[x^2+x\right]}_{-\frac{1}{2}}^1+{\left[\ln \left|x\right|\right]}_1^3\\ & =& \left(1^2+1-\left({\left(-\frac{1}{2}\right)}^2+\left(-\frac{1}{2}\right)\right)\right)+\left(\ln \left(3\right)-\ln \left(1\right)\right)\\ & =& \left(2-\left(-\frac{1}{4}\right)\right)+\left(\ln \left(3\right)-0\right)\\ & =& \frac{9}{4}+\ln 3\end{array}$

$\begin{array}{rcl}{\int }_{-\frac{1}{2}}^{3}f\left(x\right)dx& =& \frac{9}{4}+\ln 3\end{array}$