The curve is in quadrant one over the interval , which are the bounds of integration.  To see this, note that the x-intercepts are 0 and 2 and the parabola opens downward. 

The definite integral below is solved by taking the antiderivative of each term of the given polynomial function, evaluating this antiderivative at the bounds of integration, and subtracting the values.

For this particular integral use the rule,  to solve.

We can find the area under the curve by taking the anti-derivative of the function and using the two boundaries as x values.  The anti-derivative of  is .  If we use the two boundaries, we end up with our answer, .

For this integral one may be tempted to directly integrate; however, this is no rule of integrals which allows us to do so.

We must apply a complex method of integration, here u-subtitution works best.

We notice that Cos(x) is the derivative of Sin(x) so it may be best to let .

Now that we have found a proper u and du, we may directly substiute into our original integral

However, our limits are in terms of x so we must substitute our u=Sin(x) back in before evaluating

If then the graph must be concave up at the point. Based on the picture, we know that the curve is concave up on  at best. The only value that falls on this interval is , which is . Since , this definitely falls on the interval given and we can be sure it is concave up based on the picture.

To find the critical point, you must find the derivative first. To do that, multiply the exponent by the coefficient in front of the  and then subtract the exponent by . Therefore, the derivative is: . Then, to find the critical point, set the derivative equal to .

.

Inflection points can only occur when the second derivative is zero or undefined. Here we have

. 

Therefore possible inflection points occur at  and . However, to have an inflection point we must check that the sign of the second derivative is different on each side of the point. Here we have

. 

Hence, both are inflection points

Possible inflection points occur when  . This occurs at three values, . However, to be an inflection point the sign of  must be different on either side of the critical value. Hence, only  are critical points.

A point of inflection is found where the graph (or image) of a function changes concavity. To find this algebraically, we want to find where the second derivative of the function changes sign, from negative to positive, or vice-versa. So, we find the second derivative of the given function  

The first derivative using the power rule  

 is,

 and the seconds derivative is  

We then find where this second derivative equals .   when .

We then look to see if the second derivative changes signs at this point. Both graphically and algebraically, we can see that the function  does indeed change sign at, and only at, , so this is our inflection point.

In order to find the points of inflection, we need to find  using the power rule, .

Now we set , and solve for .

To verify this is a true inflection point we need to plug in a value that is less than it and a value that is greater than it into the second derivative. If there is a sign change around the  point than it is a true inflection point.

Let 

Now let 

Since the sign changes from a positive to a negative around the point , we can conclude it is an inflection point.

In order to find the points of inflection, we need to find  using the power rule .

Now to find the points of inflection, we need to set .

.

Now we can use the quadratic equation.

Recall that the quadratic equation is

,

where a,b,c refer to the coefficients of the equation  .

In this case, a=12, b=0, c=-4.

Thus the possible points of infection are

.

Now to check if or which are inflection points we need to plug in a value higher and lower than each point. If there is a sign change then the point is an inflection point.

To check  lets plug in .

Therefore  is an inflection point. 

Now lets check  with .

Therefore  is also an inflection point.