Compute the derivative for the function defined as

without evaluating the integral directly.

Let

Then, by the fundamental theorem of calculus,

So, using the linearity of the integral with respect to the interval of integration we write,

Now, differentiating and using the chain rule we have

Find the derivative for each of the following (without attempting to evaluate the integrals)

- ,
- ,
- .

For all parts of the problem let

From the fundamental theorem of calculus we then know

- Now, so
- In this case so
- Now, we rewrite the integral,
Therefore,

From above we know

Therefore,

Let be an everywhere continuous function with

Define another function by

Prove that

Also, compute the values, and .

Proof. We start with the equation for and use properties of the integral and the fundamental theorem of calculus to rearrange,

We can pull the out of the integral since it does not depend on (which is what we are taking the integral over). Then, we want to take the derivative with respect to ; however, we need to be careful and use the product rule. Looking closely at the first term we have

Where we used the fundamental theorem to obtain . So, using this method to take the derivative for we have

Now, to evaluate the second and third derivatives at 1 as requested, we first get expressions for and .

Therefore,

Let

Find a quadratic polynomial such that

Do not attempt to evaluate the integral.

First,

Hence,

Next, by the fundamental theorem of calculus,

Then, evaluate the derivative of the polynomial,

To get the final constant we take the derivative of ,

Therefore,

Hence, we have

Find a function and a constant such that

Let . Then, taking derivatives, and using the fundamental theorem of calculus we have

Then, to find the value of the constant, let and evaluate

Find a function and a constant such that

Let and . Then,

Find a function and a constant such that

Let and . Then,

Let be an everywhere continuous function satisfying

for all . Find the values of

Let

By the first fundamental theorem of calculus we know the derivative exists (since is continuous by assumption) and we have

Therefore, taking the derivative of we have,

Then, evaluating at ,

Then, taking the derivative of we have,

Therefore,