Numerical approximations to definite integrals - AP Calculus AB

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)

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

The answer is 10.

The answer is

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

f'(x)=2x+2

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

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.)

The answer is .

We know that

so,

The answer is

We simply differentiate by parts, remembering our trig rules.

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.

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}$$}

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.

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=alog x, where is a constant.

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

We can differentiate both sides with respect to .

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

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

Because we are looking for the derivative, we must solve for .

frac{dy}{dx}=ycdot 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 with .

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

First, we can write out the first few terms of the sequence , where ranges from 1 to 3.

Notice that each term , is found by multiplying the previous term by . Therefore, this sequence is a geometric sequence with a common ratio of . We can find the sum of the terms in an infinite geometric sequence, provided that , where is the common ratio between the terms. Because in this problem, is indeed less than 1. Therefore, we can use the following formula to find the sum, , of an infinite geometric series.

The answer is .

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.

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

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.

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

The answer is 2.

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:

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 = .

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

The answer is .

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

.

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)

, is the width of each subinterval, which we will determine shortly.

and 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 , 4 times; 1 time each for , , , and . This gives us

Think of 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 and are.

To find 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,

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 . In other words

Add the again to get

And repeat to find

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

plug each value into and then simplify.

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

Because our is constant, the left Riemann sum will be