AP Calculus BC

$\begin{align*}&\text{For the function: }f(x)=cos(x) +\frac{ 1}{cos(8x)}\&\text{Approximate }f(5)\&\text{Using a Taylor expansion centered at }-4\cdot \pi\text{ to }3\text{ terms}\end{align*}$

Explanation

$\begin{align*}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }cos(x) +\frac{ 1}{cos(8x)}\text{ at }x=5\&\text{This expansion becomes:}\&f(x)\approx (cos(a) +\frac{ 1}{cos(8a)})+(\frac{(8sin(8a))}{cos(8a)^{2}}- sin(a))(x-a)^{1}+\frac{\frac{64}{cos(8a)}- cos(a) +\frac{ (128sin(8a)^{2})}{cos(8a)^{3}}}{2}(x-a)^{2}\&x=5\&a=-4\cdot \pi\&f(5)\approx9722.19\end{align*}$

Find the derivative of $f(x)=5x^{2}+12x-7$ using the definition of the derivative.

Explanation

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=5x^{2}+12x-7$

$f(x+h)=5(x+h)^{2}+12(x+h)-7$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{5(x+h)^{2}+12(x+h)-7-(5x^{2}+12x-7)}{h}$

$=\lim_{h\rightarrow 0}\frac{5(x^{2}+2xh+h^{2})+12x+12h-7-5x^{2}-12x+7}{h}$

$=\lim_{h\rightarrow 0}\frac{5x^{2}+10xh+5h^{2}+12x+12h-7-5x^{2}-12x+7}{h}$

$=\lim_{h\rightarrow 0}\frac{10xh+5h^{2}+12h}{h}$

$=\lim_{h\rightarrow 0}\frac{h(10x+5h+12)}{h}$

$=\lim_{h\rightarrow 0}10x+5h+12$

Solve the integral:

$\int_{0}^{1}xe^{2x}dx$

$= \frac{1}{4}(7e^{2}+1)$

$= \frac{1}{4}(7e^{2}-1)$

$= \frac{1}{2}(7e^{2}+1)$

$= \frac{1}{2}(1-7e^{2})$

None of the chocies.

Explanation

$\int xe^{2x}dx =\int udv = uv - \int vdu$

Must choose what which term is or .

, take the derivative to get

$dv = e^{2x}dx$ , integrate to get $v = \frac{e^{2x}}{2}$

Plug into parts into the equation:
$uv - \int vdu$ $= xe^{2x} - \int \frac{1}{2} e^{2x}dx$

Integrating again, while including the bounds gives:

$\Big[e^{2x}(x-\frac{1}{4})\Big]^{1}_{0} = e^{2}(\frac{7}{4}) + \frac{1}{4}$ $= \frac{1}{4}(7e^{2}+1)$

$\begin{align*}&\text{For the function: }f(x)=-cos(6x)sin(2x)\&\text{Approximate }f(9)\&\text{Using a Taylor expansion centered at }-\frac{(3\cdot \pi )}{2}\text{ to }2\text{ terms}\end{align*}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }-cos(6x)sin(2x)\text{ at }x=9\&\text{This expansion becomes:}\&f(x)\approx (-cos(6a)sin(2a))+(6sin(2a)sin(6a) - 2cos(2a)cos(6a))(x-a)^{1}\&x=9\&a=-\frac{(3\cdot \pi )}{2}\&f(9)\approx-27.42\end{align*}$

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{0}^{6.6}(-196cos(3x)^{2})dx\&\text{Using }3\text{ intervals.}\end{align*}$

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{0}^{6.6}(-196cos(3x)^{2})dx\&\text{So the interval is }[0,6.6]\text{ the subintervals have length }\frac{6.6-(0)}{3}=\frac{11}{5}\&\text{and since we are using the *mid*point of each interval, the x-values are:}\&[\frac{11}{10},\frac{33}{10},\frac{11}{2}]\&\int_{0}^{6.6}(-196cos(3x)^{2})dx=\frac{11}{5}[(-196cos(\frac{33}{10})^{2})+(-196cos(\frac{99}{10})^{2})+(-196cos(\frac{33}{2})^{2})]\&\int_{0}^{6.6}(-196cos(3x)^{2})dx=-974.14\end{align*}$

$\begin{align*}&\text{Evaluate the indefinite integral: }\&\int(\frac{e^{(9x)}}{3}+\frac{ (64x^{10})}{7})dx\end{align*}$

$\frac{e^{(9x)}}{27}+\frac{ (64x^{11})}{77}+C$

$3e^{(9x)} +\frac{ (6400x^{11})}{77}$

$3e^{(9x)} +\frac{ (6400x^{11})}{77}+C$

$\frac{e^{(9x)}}{27}+\frac{ (64x^{11})}{77}$

Explanation

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\&\text{use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside. Integrals, in}\&\text{similar fashion, entail a division. Now, although we have two}\&\text{functions, since they're being added, we integrate separately}\&\text{and combine the results. This is known as superposition.}\&\text{Utilizing integral rules:}\&\int[e^{ax}]=\frac{e^{ax}}{a}\&\int x^{a}=\frac{x^{a+1}}{a+1}\&\text{The }a\text{ value for }\frac{e^{(9x)}}{3}\text{ is }9\&\text{The }a\text{ value for }\frac{(64x^{10})}{7}\text{ is }10\&\int(\frac{e^{(9x)}}{3}+\frac{ (64x^{10})}{7})dx=\frac{e^{(9x)}}{27}+\frac{ (64x^{11})}{77}+C\&\text{Now you'll note after integration we add a constant, C, because}\&\text{we're working with an indefinite integral.}\&\end{align*}$

$\begin{align*}&\text{Evaluate the indefinite double integral: }\&\iint(4x^{17})dx\end{align*}$

$\frac{(2x^{19})}{171}+C_{1}x+C_{2}$

$\frac{(2x^{19})}{171}$

$\frac{(2x^{18})}{9}+C_{1}x$

$\frac{(2x^{19})}{171}+C_{1}+C_{2}$

Explanation

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int x^{a}=\frac{x^{a+1}}{a+1}\&\text{Note for this problem, the }a\text{ value is } 17\&\text{Taking the first integral we find:}\&\frac{(2x^{18})}{9}+C_{1}\&\text{Note after indefinite integration we add a constant, }C_1\&\text{Integrating again, we'll add a new constant, while multiplying}\&\text{our first one by }x:\&\frac{(2x^{19})}{171}+C_{1}x+C_{2}\&\end{align*}$

Determine the following integral:

$\int (3tan(2x)sec(2x))$

$\frac{3}{2}sec(2x)$

$\frac{3}{2}csc(2x)$

$\frac{3}{2}sec(x)$

Explanation

To determine this indefinite integal, we integrate by substitution:

We can substitute for .

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}2x=2$

We can rewrite the original integral as:

$\int 3tan(2x)sec(2x) dx=\int 3tan(u)sec(u)du=$

$=\int\frac{3}{2}tan(u)sec(u)du$

Recall that $\int tan(x)sec(x) dx= sec(x)+C$ , where is a constant

Therefore:

$\int\frac{3}{2}tan(u)sec(u)du = \frac{3}{2}sec(u) +C$

Since

${3}sec(u) du +C= \frac{3}2{}sec(2x)+C$ , where is a constant

Use the method of partial fractions to evaluate the following integral:

$\int \frac{3x+11}{x^{2}-x-6}dx$

$ln(\frac{x-3}{x+2})$

$\frac{3}{2}x^{2}-5$

$-\frac{3}{2}x^{2}+5$

Explanation

When we use the method of partial fractions, our first step is to factor the polynomial in the denominator to see what the terms are going to be in the denominators of each of our partial fractions. So first we factor the denominator of the equation we're integrating, and then we write the equation in terms of our partial fractions:

$\frac{3x+11}{x^{2}-x-6}=\frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}$

If we were to add our two partial fractions together, they would need a common denominator. As with any other fractions, we find our common denominator by multiplying the numerator and denominator of one fraction by the denominator of the other. Then we add these fractions with like denominators together. In this case, we have:

$\frac{A}{x-3}+\frac{B}{x+2}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}$

This whole term is still equal to the original equation we're integrating, and now they both have the same denominator, as shown below, so we know their numerators must be equal as well:

$\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}$

Our next step is to solve for A and B. We do this by evaluating our expression above for some x value that will cancel our A term out, solving for B, and then evaluating it again for some x value that will cancel our B term out, solving for A. If we set the factors that A and B are multiplied by equal to 0, we can see that our A term will be 0 when x= -2, and our B term will be 0 when x=3. So now we evaluate our expression at these two values of x and solve for the term that is not cancelled out:

$x=-2\rightarrow 5=A(0)+B(-5)\rightarrow B=-1$