Introductory Calculus - Pre-Calculus

If we want to approximate the area under a curve using n=4, that means we will be using 4 rectangles. Because the problem asks us to approximate the area from x=0 to x=4, this means we will have a rectangle between x=0 and x=1, between x=1 and x=2, between x=2 and x=3, and between x=3 and x=4. We want to use a midpoint Riemann sum, so the height of each rectangle will be the value of the function at the midpoint of each interval:

Now that we know the height of each rectangle, all we have to do is find its area by multiplying the height by the width, which is just 1 for each rectangle. Then we add the area of all the rectangles to find our approximation for the area under the curve from x=0 to x=4:

A left hand approximation requires us to slice the region into four rectangles of equal width. Since the interval on which we are slicing is and we want to create four rectangles of equal width, the width of each rectangle must be,

.

The height of each rectangular slice is given by the function value at the left edge of each rectangle. Beginning at the leftmost edge of the first rectangle on the interval, these left endpoints exist at , but not at as that is the right edge of the fourth and final rectangle on the interval.

Using the function given, , the first rectangle has height . The rest have height .

Multiply each by the width of each rectangle to get the area of each rectangular slice. Then add the area of the slices together, and simplify to get the correct result.

The interval divided into four sub-intervals gives rectangles with vertices of the bases at

For the approximated area, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or , , , and

Because each sub-interval has width , the approximated area using rectangles is

As such, the approximated area is

units squared

The interval divided into two sub-intervals gives two rectangles with vertices of the bases at

For the approximated area, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or and

Because each sub-interval has width , the approximated area using rectangles is

As such, the approximated area is

units squared

In order to approximate the area under a curve using rectangles, one must take the sum of the areas of discrete rectangles under the curve. Taking the height of each rectangle as the function evaluated at the left endpoint, we obtain the following rectangle areas:

The sum of the individual rectangles yields an overall area approximation of 100.

In order to approximate the area under a curve using rectangles, one must take the sum of the areas of discrete rectangles under the curve. Taking the height of each rectangle as the function evaluated at the right endpoint, we obtain the following rectangle areas:

The sum of the individual rectangles yields an overall area approximation of 225.

The area under the curve of the function is the definite integral from to .

Remember when integrating, we will increase the exponent by one and then divide the whole term by the value of the new exponent.

From here, we find the difference between the function values of the boundaries.

First find the slope of the tangent to the line by taking the derivative.

Using the Exponential Rule we get the following,

.

Then plug 1 into the equation as 1 is the point to find the slope at.

.

One way of finding the slope at a given point is by finding the derivative. In this case, we can take the derivative of y with respect to x, and plug in the desired value for x.

Using the exponential rule we get the following derivative,

.

Plugging in x=2 from the point 2,3 gives us the final slope,

Thus our slope at the specific point is .

Note that in this case, using the y coordinate was not necessary.

To find the slope of the tangent line of the function at the given value, evaluate the first derivative for the given.

The first derivative is

and for this function

and

So the slope is

To find the slope of the tangent line of the function at the given value, evaluate the first derivative for the given value.

The first derivative is

and for this function

and plugging in the specific x value we get,

So the slope is

.

Calculate the derivative of by using the derivative rules. The derivative function determines the slope at any point of the original function.

The derivative is:

With the given point , . Substitute this value to the derivative function to determine the slope at that point.

The slope of the tangent line that intersects point is .

We begin by recalling that one way of defining the derivative of a function is the slope of the tangent line of the function at a given point. Therefore, finding the derivative of our equation will allow us to find the slope of the tangent line. Since the two things needed to find the equation of a line are the slope and a point, we would be halfway done.

We calculate the derivative using the power rule.

However, we don't want the slope of the tangent line at just any point but rather specifically at the point . To obtain this, we simply substitute our x-value 1 into the derivative.

Therefore, the slope of our tangent line is .

We now need a point on our tangent line. Our choices are quite limited, as the only point on the tangent line that we know is the point where it intersects our original graph, namely the point .

Therefore, we can plug these coordinates along with our slope into the general point-slope form to find the equation.

Solving for will give us our slope-intercept form.

The equation of the tangent line at depends on the derivative at that point and the function value.

The derivative at that point of is

using the Power Rule

which means

The derivative is zero, so the tangent line will be horizontal.

It intersects it at since , so that line is .

Rewrite in slope-intercept form, , to determine the slope.

The slope of the given function is 2.

Substitute the slope and the given point, , in the slope-intercept form to determine the y-intercept.

Substitute this and the slope back to the slope-intercept equation.

The equation of the tangent line is:

We begin by finding the equation of the derivative using the limit definition:

We define and as follows:

We can then define their difference:

Then, we divide by h to prepare to take the limit:

Then, the limit will give us the equation of the derivative.

Now, we must realize that the slope of the line tangent to the curve at the given point is equivalent to the derivative at the point. So if we define our tangent line as: , then this m is defined thus:

Therefore, the equation of the line tangent to the curve at the given point is:

First, find the slope of this tangent line by taking the derivative:

Plugging in 1 for x:

So the slope is 4

Now we need to find the y-coordinate when x is 1, so plug 1 in to the original equation:

To write the equation, use point-slope form and then use algebra to change to slope-intercept like the answer choices:

distribute the 4

add 2 to both sides

First, find the slope of the tangent line by taking the first derivative:

To finish determining the slope, plug in the x-value, 2:

the slope is 6

Now find the y-coordinate where x is 2 by plugging in 2 to the original equation:

To write the equation, start in point-slope form and then use algebra to get it into slope-intercept like the answer choices.

distribute the 6

add 8 to both sides

First, take the first derivative in order to find the slope:

To continue finding the slope, plug in the x-value, -2:

Then find the y-coordinate by plugging -2 into the original equation:

The y-coordinate is

Now write the equation in point-slope form then algebraically manipulate it to match one of the slope-intercept forms of the answer choices.

distribute the -5

add to both sides

First distribute the . That will make it easier to take the derivative:

Now take the derivative of the equation:

To find the slope, plug in the x-value -3:

To find the y-coordinate of the point, plug in the x-value into the original equation:

Now write the equation in point-slope, then use algebra to get it into slope-intercept like the answer choices:

distribute

subtract from both sides

write as a mixed number