The definite integral  can be integrated as is:

Evaluate the bounds.  Be aware of the sign changes.

Be careful, we are integrating with the respect to y, not x.  This means that the integrand itself is treated as a constant.

Since we just have a constant, to integrate means to increase the value of the variable y. Thus we get the following,

.

From here we plug in our bounds. We take the upper bound function value and subtract lower bound function value.

To help us evalute the integral, we can split up the expression into 3 parts:

.

This allows us to evaluate the integral of each of the three parts, sum them up, and then evaluate the summed up parts from 0 to 1.

The first integral is .

The second integral is .

The third integral is .

Sum up all these terms in evaluate between 0 and 1.

The general expression for the integral of a sine function is:

Using this, the integral above now becomes,

, which simplifies to

To integrate, we must first make the following subsitution:

The derivative was found using the following rules:

, 

Now, rewrite the given integral, change the bounds in terms of  (by plugging in the upper and lower bounds into the equation for  in terms of ), and integrate:

The integral was performed using the following rule:

To perform the definite integration, simply plug in the upper limit of integration and subtract from the result of plugging in the lower limit of integration, as shown above. Simplifying the result, we get 

In order to calculate the definite integral, we apply the inverse power rule which states

Applying this to the problem in this question term by term we get

And by the corollary of the Fundamental Theorem of Calculus the definite integral becomes

And so

In order to calculate the definite integral, we apply the inverse power rule which states

Applying this to the problem in this question we get

And by the corollary of the Fundamental Theorem of Calculus the definite integral becomes

And so

The integral can be rewritten as:

The integration of a constant is simply the constant times the variable it is integrating with respect to.  Plug in the upper bound and subtract after substituting the lower bound.

There is no need to add a  term at the end when we are dealing with definite integrals.

The answer is:  

Integrate each term.

Substitute the upper bound and subtract the quantity after substituting of the lower bound.

The right hand side of the equation becomes:

Convert all fraction to the least common denominator.

Simplify the terms.

The answer is:  

Must choose what which term is  or .

, take the derivative to get  

, integrate to get 

Plug into parts into the equation:
  

Integrating again, while including the bounds gives: