A Riemann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

We're asked to approximate 

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are 

A Riemann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

We're asked to approximate 

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are 

A Riemann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

We're asked to approximate 

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are 

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

For the Midpoint Riemann sum, we need to find the rectangle heights which come from the midpoint of the sub-intervals, or , , and .

Because each sub-interval has a width of , the Midpoint Riemann sum is

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

For the Midpoint Riemann sum, we need to find the rectangle heights which come from the midpoint of the sub-intervals, or , , , and .

Because each sub-interval has a width of , the Midpoint Riemann sum is

Riemann Sums are a way to approximate the area under a curve. While taking an integral gives you an exact area, using Riemann Sums is a way to get a fast estimation given a function that might be more difficult to integrate. There are many different ways to use this technique. Given some number, n (which as it increases gives you better approximations), we can split the area under the curve into n rectangles. The height of each rectangle depends on what version of Riemann Sums you are using. Three such versions are left endpoint, right endpoint, and midpoint. The formula itself is simple, say we have a curve bounded in [a,b]. Then we can approximate the area using the following equation 

.

Where the  are a list of midpoints for each rectangle and  which is the width of each rectangle. 

First we calculate the length of each subinterval 

.  

Now we look at the first rectangle.  

It is in between 5 and 7 so the midpoint of the two is 6.  

Now we easily know the midpoint of all 5 of the rectangles 

.  

Thus the area is thus 

which is simply 

.

To find the derivative, we must use both the product and the chain rule.  Let us consider each part:

f(x) = sin(x²)

g(x) = 3x² + 5x

For f(x), we will need to apply the chain rule.  Consider f(x) as being f(h(x)) where h(x) = x^2; therefore f'(x) = f'(h(x)) * h'(x) or cos(x²) * 2x = 2x * cos(x²).

Gathering together all of our data, we have:

f(x) = sin(x²)

f'(x) = 2x * cos(x²)

g(x) = 3x² + 5x

g'(x) = 6x + 5

The product rule states that if a(x) = f(x) * g(x), then a'(x) = f'(x) * g(x) + f(x) * g'(x)

For our data, then, the derivative of a(x) = sin(x²)(3x² + 5x) is:

a'(x) = 2x * cos(x²) * (3x² + 5x) + sin(x²) * (6x + 5)

Distributing, this gives us:

a'(x) = 6x³cos(x²) + 10x²cos(x²) + 6x * sin(x²)+ 5sin(x²)

For each of the parts of f(x), you will have to apply the chain rule:

For sin(2x² + 5x): cos(2x² + 5x) * (4x + 5)

For (sin(cos(x)): cos(cos(x))*(-sin(x))

Therefore, for our original:

f'(x) = cos(2x² + 5x) * (4x + 5) – cos(cos(x)) * (–sin(x))

Simplifying:

f'(x) = 4x*cos(2x² + 5x) + 5cos(2x² + 5x) + cos(cos(x)) * sin(x)

f(x) = ln(x^{2 *} tan(x²))?

The chain rule is necessary.  For the natural logarithm "portion", we know the derivative will be:1/(x² * tan(x²))

However, we must add the derivative of the argument using the product rule: 2x*tan(x²) + x² * 2x * sec²(x²)

The whole derivative: f'(x) = (2x * tan(x²) + x² * 2x * sec²(x²))/(x² * tan(x²))

Splitting apart the fraction, we have the sum of:

(2x * tan(x²))/(x² * tan(x²))

and

x² * 2x * sec²(x²)/(x² * tan(x²))

The first simplifies to 2/x

The second is a bit trickier. First let us simplify the non-trigonometric components:

2x * sec²(x²)/tan(x²)

Now, since sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x), we can rewrite:

2x * sec²(x²)/tan(x²) = 2x/(cos²(x²) * sin(x²)/cos(x²))

This reduces to: 2x/(cos(x²) * sin(x²)), which is the same as 2x * sec(x²) * csc(x²)

Therefore, we can write our solution as f'(x) = (2/x) + 2x * sec(x²) * csc(x²)

We must carefully parse this function in order to apply the chain rule correctly. Let us make the following substitutions:

f(x) = 2^x

g(x) = sin(sin(x))

s(x) = f(g(x)) = 2^sin(sin(x))

According to the chain rule, we know that s'(x)  = f'(g(x)) * g'(x)

f'(g(x)) = ln(2) * 2^sin(sin(x))

g'(x) = cos(sin(x)) * cos(x)

Therefore, s'(x) = ln(2) * 2^sin(sin(x))  * cos(sin(x)) * cos(x)