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AP Calculus BC › Integrals

Questions 1 - 10

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\&\int_{-4}^{10}(-14ln(2x^{2}))dx\&\text{Using }4\text{ intervals.}\end{align*}$

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-4}^{10}(-14ln(2x^{2}))dx\&\text{So the interval is }[-4,10]\text{ the subintervals have length }\frac{10-(-4)}{4}=\frac{7}{2}\&\text{and since we are using the *right*point of each interval, the x-values are:}\&[-\frac{1}{2},3,\frac{13}{2},10]\&\int_{-4}^{10}(-14ln(2x^{2}))dx=\frac{7}{2}[(14ln(2))+(-14ln(18))+(-14ln(\frac{169}{2}))+(-14ln(200))]\&\int_{-4}^{10}(-14ln(2x^{2}))dx=-584.68\end{align*}$

he Laplace Transform is an integral transform that converts functions from the time domain to the complex frequency domain . The transformation of a function into its Laplace Transform is given by:

$F(s)= \int_{0}^{\infty}e^{-st}*f(t) dt$

Where , where and are constants and is the imaginary number.

Give the Laplace Transform of $\frac{\mathrm{d} }{\mathrm{d} t}[f(t)]$ .

Explanation

The Laplace Transform of the derivative is given by:

$\int_{0}^{\infty}e^{-st}*f'(t) dt$

Using integration by parts,

$\int_{0}^{\infty} UdV= [UV]_{0}^{\infty}-\int_{0}^{\infty} VdU$

Let and $U=e^{-st}$

$[e^{-st}*f(t) ]_{0}^{\infty}-\int_{0}^{\infty}(-s)e^{-st}*f(t) dt$

The first term becomes

and the second term becomes

The Laplace Transform therefore becomes:

Given a function $y=-6x^{2}$ , find the Left Riemann Sum of the function on the interval divided into three sub-intervals.

Explanation

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval divided into sub-intervals, we'll be using rectangles with vertices at .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is because the rectangles are spaced units apart. Since we're looking for the Left Riemann Sum, we want to find the heights of each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

$f(0)=-6(0)^{2}=0$

$f(2)=-6(2)^{2}=-24$

$f(4)=-6(4)^{2}=-96$

Putting it all together, the Left Riemann Sum is

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{-5}^{6.4}(4\cdot 3^x)dx\&\text{Using }3\text{ intervals.}\end{align*}$

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-5}^{6.4}(4\cdot 3^x)dx\&\text{So the interval is}[-5,6.4]\text{ the subintervals have length}\frac{6.4-(-5)}{3}=3.8\&\text{and since we are using the *mid*point of each interval, the x-values are:}\&[-\frac{31}{10},\frac{7}{10},\frac{9}{2}]\&\int_{-5}^{6.4}(4\cdot 3^x)dx=3.8[(\frac{(4\cdot 3^{(\frac{9}{10})})}{81})+(4\cdot 3^{(\frac{7}{10})})+(324\cdot 3^{(\frac{1}{2})})]\&\int_{-5}^{6.4}(4\cdot 3^x)dx=2165.80\end{align*}$

Evaluate.

$\int 3x^2 + 12x^{11} \ dx$

$F(x) = x^3 + x^{12}$

$F(x) = 3x^3 + 12x^{12}$

$F(x) = \frac{1}{3}x^3 +\frac{1}{12} x^{12}$

$F(x) = x^2 + x^{11}$

Answer not listed

Explanation

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of

If then

If then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then

If then

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = 3x^2 + 12x^{11}$ .

The antiderivative is $F(x) = x^3 + x^{12}$ .

Evaluate.

$\int 3 \ dx$

$F(x) = \frac{1}{3}x+ C$

Answer not listed

Explanation

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of

If then

If then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then

If then

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, .

The antiderivative is .

Evaluate.

$\int 12 + x \ dx$

$F(x) = 12x + \frac{x^2}{2} + C$

$F(x) = \frac{6x}{2} + \frac{x^2}{2} +C$

Answer not listed.

Explanation

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of

If then

If then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then

If then

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, .

The antiderivative is $F(x) = 12x + \frac{x^2}{2} +C$ .

Given a function $y=-6x^{2}$ , find the Left Riemann Sum of the function on the interval divided into three sub-intervals.

Explanation

$f(0)=-6(0)^{2}=0$

$f(2)=-6(2)^{2}=-24$

$f(4)=-6(4)^{2}=-96$

Putting it all together, the Left Riemann Sum is

Evaluate.

$\int x^3 - \cos (2x) \ dx$

$F(x) = \frac{x^4}{4} - \frac{\sin (2x)}{2}+C$

$F(x) = x^4 - \sin (2x)$

$F(x) = \frac{x^3}{4} - \frac{\sin (x)}{2}+C$

$F(x) = x^4 + \frac{\sin (2x)}{2}$

Answer not listed

Explanation

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of

If then

If then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then

If then

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = x^3 - \cos (2x)$ .

The antiderivative is $F(x) = \frac{x^4}{4} - \frac{\sin (2x)}{2}+C$ .

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

Explanation

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

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