Numerical approximations to definite integrals - AP Calculus AB

Card 0 of 990

Question

Write the equation of a tangent line to the given function at the point.

y = ln(x2) at (e, 3)

Answer

To solve this, first find the derivative of the function (otherwise known as the slope).

y = ln(x2)

y' = (2x/(x2))

Then, to find the slope in respect to the given points (e, 3), plug in e.

y' = (2e)/(e2)

Simplify.

y'=(2/e)

The question asks to find the tangent line to the function at (e, 3), so use the point-slope formula and the points (e, 3).

y – 3 = (2/e)(x – e)

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Question

If $f(x)=4, f'(x)=3, g(x)=2, \ and\ g'(x)=1$ then find .

$h(x)=\frac{2f}{g}$

Answer

The answer is 1.

$h(x)=\frac{2f}{g}$

$h'(x)=\frac{2(f'(x)g(x)-f(x)g'(x))}{g^{2}}$

$h'(x)=\frac{2(32-41)}{4} = 1$

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Question

If $f(x)=1, f'(x)=2, g(x)=3, \ and\ g'(x)=4$ then find .

Answer

The answer is 10.

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Question

Find the equation of the tangent line at on graph

Answer

The answer is

$f(x)=x^{2}+2x-8$

$f'(x)=2x+2$

(This is the slope. Now use the point-slope formula)

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Question

Find the equation of the tangent line at (1,1) in

$f(x)=ax^{2}+bx+c$

Answer

The answer is

$f(x)=ax^{2}+bx+c$

$f'(x)=2ax+b$

$f'(1)=2a(1)+b = 2a+b$

(This is the slope. Now use the point-slope formula.)

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Question

If $f(x)=\cos x$ then

$f^{'}\left ( \frac{\pi }{2} \right )=$

Answer

The answer is .

We know that

$\frac{\pi }{2}=90^{\circ}$

so,

$f'\left ( \frac{\pi }{2}\right )=-sin(90^{\circ})=-1$

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Question

Differentiate $y=3\csc x+5\sin x$

Answer

The answer is $-3\csc x\cot x+5\cos x$

We simply differentiate by parts, remembering our trig rules.

$y=3\csc x+5\sin x$

$y^{'}= -3\csc x\cot x+5\cos x$

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Question

Find the equation of the tangent line at when

$y=\frac{x^{3}-2x(x^{1/2})}{x}$

Answer

The answer is

$y=\frac{x^{3}-2x(x^{1/2})}{x}$ let's go ahead and cancel out the 's. This will simplify things.

$y=x^{2}-2(x^{1/2})$

$y'=2x-\frac{1}{(x^{1/2})}$

$y'(1)=2(1)-\frac{1}{(1^{1/2})} =1$ this is the slope so let's use the point slope formula.

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Question

Differentiate $y=\frac{t+2}{t^{2}-4t-12}$

Answer

We see the answer is $y'=\frac{-1}{(t-6)^{2}}$ after we simplify and use the quotient rule.

$y=\frac{t+2}{t^{2}-4t-12}$ we could use the quotient rule immediatly but it is easier if we simplify first.

$y=\frac{t+2}{(t+2)(t-6)}$

$y=\frac{1}{(t-6)}$

$y=(t-6)^{-1}$

$y'=-(t-6)^{-2}$

$y'=\frac{-1}{(t-6)^{2}}$

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Question

Find $\lim_{x\rightarrow \infty }\frac{-2x^3+x^2-2}{x^2+10}$

Answer

When taking limits to infinity, we usually only consider the highest exponents. In this case, the numerator has $-2x^3$ and the denominator has $x^2$ . Therefore, by cancellation, it becomes $-2x$ as approaches infinity. So the answer is $-\infty$ .

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Question

What is the first derivative of the function $\large h(x)=x^{\frac{1}{x}}$ ?

Answer

First, let .

$y=x^{\frac{1}{x}}$

We will take the natural logarithm of both sides in order to simplify the exponential expression on the right.

$\ln y=\ln x^{\frac{1}{x}}$

Next, apply the property of logarithms which states that, in general, $\log x^a=a\log x$ , where is a constant.

$\ln y = \frac{1}{x}\ln x$

We can differentiate both sides with respect to .

$\frac{d}{dx}\[ln y]=\frac{d}{dx}[\frac{1}{x}\ln x]$

We will need to apply the Chain Rule on the left side and the Product Rule on the right side.

$\frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{d}{dx}[\ln x]+\ln x\cdot \frac{d}{dx}[\frac{1}{x}]$

$\frac{1}{y}\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{1}{x} + \ln x\cdot \frac{-1}{x^{2}}$

Because we are looking for the derivative, we must solve for $\frac{\mathrm{d}y }{\mathrm{d} x}$ .

$\frac{dy}{dx}=y\cdot \frac{1}{x^{2}}(1-\ln x)$

However, we want our answer to be in terms of only. We now substitute $x^{\frac{1}{x}}$ in place of .

$\frac{dy}{dx}=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$

Since we let , we can replace $\frac{\mathrm{d} y}{\mathrm{d} x}$ with .

The answer is $h'(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$ .

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Question

Evaluate:

$\sum_{n=1}^{\infty }4(\frac{-1}{2})^{n-1}$

Answer

First, we can write out the first few terms of the sequence $a_{n}=$ $4(\frac{-1}{2})^{n-1}$ , where ranges from 1 to 3.

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

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

$a_{3}=4(\frac{-1}{2})^{3-1}=1$

Notice that each term $\left ( n>1 \right )$ , is found by multiplying the previous term by $-\frac{1}{2}$ . Therefore, this sequence is a geometric sequence with a common ratio of $-\frac{1}{2}$ . We can find the sum of the terms in an infinite geometric sequence, provided that , where is the common ratio between the terms. Because $r=-\frac{1}{2}$ in this problem, $\left | r \right |$ is indeed less than 1. Therefore, we can use the following formula to find the sum, , of an infinite geometric series.

$S=\frac{a_{1}}{1-r}=\frac{4}{1-(\frac{-1}{2})}=\frac{4}{3/2}=\frac{8}{3}$

The answer is $\frac{8}{3}$ .

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Question

$\lim_{x\rightarrow 0}\frac{2\sin(x^2)}{x\tan(x)}=$

Answer

When we let x = 0 in our original limit, we obtain the 0/0 indeterminate form. Therefore, we can apply L'Hospital's Rule, which requires that we take the derivative of the numerator and denominator separately.

$\lim_{x\rightarrow 0}\frac{2\sin(x^2)}{x\tan(x)}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}2\sin(x^2)}{\frac{\mathrm{d} }{\mathrm{d} x}x\tan(x)}$

Apply the Chain Rule in the numerator and the Product Rule in the denominator.

$\lim_{x\rightarrow 0}\frac{4x\cos(x^2)}{\tan(x)+x\sec^2(x)}$

If we again substitute x = 0, we still obtain the 0/0 indeterminate form. Thus, we can apply L'Hospital's Rule one more time.

$\lim_{x\rightarrow 0}\frac{4x\cos(x^2)}{\tan(x)+x\sec^2(x)}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}4x\cos(x^2)}{\frac{\mathrm{d} }{\mathrm{d} x}(\tan(x)+x\sec^2(x))}$

$=\lim_{x\rightarrow 0}\frac{4\cos(x^2)+4x(2x)(-\sin(x^2))}{\sec^2(x)+(\sec^2(x)+x(2\sec^2x\tan x))}$

If we now let x = 0, we can evaluate the limit.

$=\lim_{x\rightarrow 0}\frac{4\cos(x^2)+4x(2x)(-\sin(x^2))}{\sec^2(x)+(\sec^2(x)+x(2\sec^2x\tan x))}=\frac{4\cos(0)+0}{\sec^2(0)+\sec^2(0)+0}$

$=\frac{4}{1+1}=2$

The answer is 2.

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Question

Consider the curve given by the parametric equations below:

$x(t)=\ln(t)$

What is the equation of the line normal to the curve when ?

Answer

In order to find the equation of the normal line, we will need the slope of the line and a point through which it passes. If we substitute into our parametric equations, we can easily obtain the point on the curve.

The normal line is perpendicular to the tangent line. Thus, we should first find the slope of the tangent line.

To find the value of the tangent slope when , we will use the following formula:

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}$

$y'(t)=3^t \ln 3 -2^t \ln 2$

$y'(1)=3\ln3-2\ln 2=\ln \frac{27}{4}$

$x'(t)=\frac{1}{t}$

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\ln(\frac{27}{4})$

Because the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus,

slope of normal = $\frac{-1}{\ln(\frac{27}{4})}$ .

We now have the point and slope of the normal line, so we can use point-slope form.

$y-1=\frac{-1}{\ln \frac{27}{4}}(x-0)$

The answer is $y=\frac{-1}{\ln \frac{27}{4}}x+1$ .

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Question

Use Left Riemann sums with 4 subintervals to approximate the area between the x-axis, , , and .

Answer

To use left Riemann sums, we need to use the following formula:

$Area \approx \sum_{i=1}^n f(x_i)\Delta x$ .

where is the number of subintervals, (4 in our problem),

is the "counter" that denotes which subinterval we are working with,(4 subintervals mean that will be 1, 2, 3, and then 4)

is the function value when you plug in the "i-th" x value, (i-th in this case will be 1-st, 2-nd, 3-rd, and 4-th)

, $\Delta x$ is the width of each subinterval, which we will determine shortly.

and $\sum_{i=1}^n$ means add all versions together (for us that means add up 4 versions).

This fancy equation approximates using boxes. We can rewrite this fancy equation by writing $f(x_i)\Delta x$ , 4 times; 1 time each for , , , and . This gives us

$f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x$

Think of $\Delta x$ as the base of each box, and as the height of the 1st box.

This is basically , 4 times, and then added together.

Now we need to determine what $\Delta x, x_1, x_2, x_3,$ and are.

To find $\Delta x$ we find the total length between the beginning and ending x values, which are given in the problem as and . We then split this total length into 4 pieces, since we are told to use 4 subintervals.

In short, $\Delta x = \frac{b - a}{n}= \frac{(2)- (-2)}{4} = 1$

Now we need to find the x values that are the left endpoints of each of the 4 subintervals. Left endpoints because we are doing Left Riemann sums.

The left most x value happens to be the smaller of the overall endpoints given in the question. In other words, since we only care about the area from to , we'll just use the smaller one, , for our first .

Now we know that .

To find the next endpoint, , just increase the first x by the length of the subinterval, which is $\Delta x = 1$ . In other words

$x_2 = x_1 + \Delta x$

Add the $\Delta x$ again to get

$x_3 = x_2 + \Delta x$

And repeat to find

$x_4 = x_3 + \Delta x$

Now that we have all the pieces, we can plug them in.

$f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x$

$f(-2)\cdot1 + f(-1) \cdot (1)+ f(0)\cdot (1) + f(1)\cdot (1)$

plug each value into and then simplify.

This is the final answer, which is an approximation of the area under the function.

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Question

Solve using the trapezoidal approximation:

$\int_{0}^{2}(x^5+e^x)dx$

Answer

The trapezoidal approximation of a definite integral is given by the following:

$\int_{a}^{b}f(x)dx \approx \frac{f(b)+f(a)}{2}(b-a)$

Using this approximation for our integral, we get

$(2-0)(\frac{32+e^2+1}{2})=33+e^2$

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Question

Solve the integral using the trapezoidal approximation:

$\int_{0}^{4}xe^{2x}dx$

Answer

To approximate the definite integral using the trapezoidal rule, we use the following approximation:

$\int_{a}^{b}f(x)dx\approx (b-a)(\frac{f(b)+f(a)}{2})$

For our integral, we get

$\int_{0}^{4}xe^{2x}dx \approx (4-0)(\frac{4e^8}{2}+0)=8e^8$

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Question

Estimate the integral of from 0 to 3 using left-Riemann sum and 6 rectangles. Use $\Delta x=0.5$

Answer

Because our $\Delta x$ is constant, the left Riemann sum will be

$R_{L}=\Delta x(f(0)+f(0.5)+f(1)+f(1.5)+f(2)+f(2.5))$

$R_{L}=0.5\cdot (1+1.75+4+7.75+13+19.75)=23.625$

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Question

$\begin{align}&\text{The function }f(x)\text{ has the following values on the interval }x=[-2,13]:\&f(-2)=5\&f(1)=-19\&f(4)=5\&f(7)=-6\&f(10)=-18\&f(13)=0\&\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-2)=5\&f(1)=-19\&f(4)=5\&f(7)=-6\&f(10)=-18\&f(13)=0\&\&\text{Although the total interval has a length of }15\&\text{Notice the points are equidistant. This distance is our subinterval }3\&\text{and since we are using the right point of each interval, we disregard the first function value:}\&\int_{-2}^{13}f(x)\approx3[(-19)+(5)+(-6)+(-18)+(0)]\&\int_{-2}^{13}f(x)\approx-114\end{align}$

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Question

$\begin{align}&\text{Using the method of trapezoidal sums approximate the integral:}\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\&\text{Using }3\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A trapezoidal integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\&\text{It is essentially a sum of n trapezoids, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\&\text{We're asked to approximate}\int_{2}^{6.5}(17cos(2sin(2x)))dx\&\text{So the interval is }[2,6.5]\text{ the subintervals have length }\frac{6.5-(2)}{3}=\frac{3}{2}\&\text{and the x-values are:}\&[2,\frac{7}{2},5,\frac{13}{2}]\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx\frac{3}{4}[(17cos(2sin(2(2))))+2(17cos(2sin(2(\frac{7}{2}))))+2(17cos(2sin(2(5))))+(17cos(2sin(2(\frac{13}{2}))))]\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx27.55\end{align}$

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