Because both functions are in terms of \(\displaystyle x\), and because the bounds are provided in terms of \(\displaystyle x\), it is most likely that the area equation needed to solve this problem is \(\displaystyle \int_{a}^{b}(upper function)-(lower function) dx\). 

To create the correct integral, the first step is to recognize the bounds. Essentially, we want to find the area between the two functions within the domain \(\displaystyle 0\leq x\leq 2\). The next step is to determine which function sits higher along the y axis for the given domain; in this case, the “upper function” is \(\displaystyle y=x^2\). From here, values can be substituted into the general equation and the area can be solved for:

\(\displaystyle Area = \int_{0}^{2}(x^2)-(x-1) dx\)\(\displaystyle = [\frac{1}{3}x^3-\frac{1}{2}x^2+x] |_0^2 = \frac{1}{3}(2)^3-\frac{1}{2}(2)^2+2-0+0-0 = \frac{8}{3}\)

As with other area questions, the general equation used is \(\displaystyle \int_{a}^{b}(upper function)-(lower function) dx\).

This question can be challenging because there are in fact multiple regions with finite area within the domain \(\displaystyle 0\leq x\leq \frac{\pi}{2}\) created by the two functions \(\displaystyle y=cos(x\)) and \(\displaystyle y=sin(x)\). If the graph is examined closely, it can be seen that when moving from left to right, \(\displaystyle y=cos(x)\) is at first the “upper function” but later becomes the “lower function.” Therefore, we will need to write two expressions to encompass the observed behavior:

\(\displaystyle Area = \int_{0}^{\frac{\pi}{4}}cos(x)-sin(x) dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} sin(x)-cos(x) dx\)\(\displaystyle = [sin(x)+cos(x)] |_0^{\frac{\pi}{4}} + [-cos(x)-sin(x)] |_{\frac{\pi}{4}}^{\frac{\pi}{2}}\)

\(\displaystyle Area =\)\(\displaystyle sin(\frac{\pi}{4})+cos(\frac{\pi}{4})-sin(0)-cos(0)-cos(\frac{\pi}{2})-sin(\frac{\pi}{2})+cos(\frac{\pi}{4})+sin(\frac{\pi}{4})\)

\(\displaystyle Area =\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}-0-1-0-1+\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}=2\frac{\sqrt{2}}{2}-2\)

When beginning an area between curves problem, it is important to first understand what the region in question looks like. In this case, there is an intersection point between the two functions, \(\displaystyle y=2x^3\) and \(\displaystyle y=x+1\), at \(\displaystyle (1,2)\). The \(\displaystyle x\) coordinate of this intersection point will act as the upper bound for the definite integral expression. The lower bound is obtained from the problem itself; this should be the value \(\displaystyle x=0\) since the \(\displaystyle y\) axis is a boundary. 

We start with the general expression, then substitute values and solve:

\(\displaystyle \int_{a}^{b}(upper function)-(lower function) dx\)

\(\displaystyle Area = \int_{0}^{1}(x+1)-(2x^3) dx = [\frac{1}{2}x^2+x-\frac{1}{2}x^4] |_0^1 = \frac{1}{2}+1-\frac{1}{2}-0-0+0=1\)

The \(\displaystyle ln(x)\) function can be challenging to work with. In this case, when determining the boundaries, it will be much easier to consider the region in terms of \(\displaystyle y\). This means that the bounds of the definite integral expression will be \(\displaystyle 0\leq y\leq 4\), but it also means the equation \(\displaystyle y=ln(5x)\) must be rewritten in terms of \(\displaystyle y\).

\(\displaystyle y=ln(5x) \Rightarrow x=\frac{1}{5}e^y\)

Now, with \(\displaystyle x=\frac{1}{5}e^y\) as the “right function” and \(\displaystyle x=0\) as the “left function,” values can be substituted into the general formula:

\(\displaystyle Area = \int_{a}^{b}(right function)-(left function) dy\)

\(\displaystyle \int_{0}^{4}\frac{1}{5}e^y-0 dy=\int_{0}^{4}\frac{1}{5}e^y dy\)

\(\displaystyle Area=\int_{a}^{b}(upper function)-(lower function) dx\)

First, to correctly set up the definite integral, the bounds must be identified. In this case, the two functions intersect in two locations, thus providing the bounds.

To find the intersection points, set the equations equal to one another:

\(\displaystyle 4x^2+1=2x+3 \rightarrow 4x^2-2x-2=0 \rightarrow x={-0.5, 1}\)

After establishing the domain as \(\displaystyle -0.5\leq x\leq 1\), the function \(\displaystyle y=2x+3\) is found to be the “upper function,” either by graphing the functions or noticing that  \(\displaystyle 2x+3\geq 4x^2+1\) for this domain. From here, it is a matter of substituting the correct values to find the area expression:

\(\displaystyle Area =\int_{a}^{b}(upper function)-(lower function) dx\)\(\displaystyle = \int_{-0.5}^{1}(2x+3)-(4x^2+1) dx\)

\(\displaystyle Area = \int_{-0.5}^{1}-4x^2+2x+2 dx\)

This region is bounded by three sides: \(\displaystyle y=e^x\) on the top and left, \(\displaystyle y=1\) on the bottom, and \(\displaystyle x=3\) on the right. The lower bound for the definite integral expression is the intersection of \(\displaystyle y=e^x\) and \(\displaystyle y=1\), or the point \(\displaystyle (0,1)\). Because the integral will be in terms of \(\displaystyle x\), the coordinate of interest is \(\displaystyle x=0\). The upper bound is simply \(\displaystyle x=3\), as that is one of the boundaries enclosing the region.

From here, all the necessary information for the general area equation \(\displaystyle Area =\int_{a}^{b}(upper function)-(lower function) dx\) has been obtained. 

\(\displaystyle Area =\int_{0}^{3}(e^x)-(1) dx\)

This is one of the few times when the \(\displaystyle \int_{a}^{b}(right function)-(left function) dy\) general area formula is the only correct approach. This general equation differs from what is “usually” seen, because it is in terms of y and emphasizes the relation the functions have to the y axis. The  reason the \(\displaystyle \int_{a}^{b}(upper function)-(lower function) dx\) formula cannot be used is because there are sections of each function where there are two \(\displaystyle y\) values for the same \(\displaystyle x\) value. 

In other words, there really isn’t a clear “upper function” or “lower function” for this problem. Instead, the clearest approach is to use the y axis as a point of reference. Because the functions are already in terms of \(\displaystyle y\), there is no need to rewrite the formulas. There are two intersection points for these functions, creating the boundaries for the integral. Since our definite integral is in terms of \(\displaystyle y\), the y coordinate values of the intersection points are used.

\(\displaystyle \int_{a}^{b}(right function)-(left function) dy\)

\(\displaystyle Area = \int_{-1}^{3}(10-y^2)-(y-2)^2 dy =\int_{-1}^{3}10-y^2-(y2-4y+4) dy\)

\(\displaystyle Area = \int_{-1}^{3}6-2y^2+4y dy =[6x-\frac{2}{3}y^3+2y^2] |_{-1}^3\)\(\displaystyle =6(3)-\frac{2}{3}(3)^3+2(3)^2-6(-1)+\frac{2}{3}(-1)^3-2(-1)^2\)

\(\displaystyle Area = 18-18+18+6-\frac{2}{3}-2 =\frac{64}{3}\)

\(\displaystyle a(t)=(a+b)t\)

Begin by finding the velocity function by integrating the acceleration function.

 \(\displaystyle v(t)=\int a(t)dt = \int (a+b)tdt=\frac{1}{2}(a+b)t^2+v_o\) 

We use \(\displaystyle v_o\) as the constant of integration since the function \(\displaystyle v(t)\) is a velocity and at initially, at \(\displaystyle t = 0\), the velocity is equals the constant of integration. 

 \(\displaystyle v(t)=\frac{1}{2}(a+b)t^2+v_o\)

At \(\displaystyle t = 10\) seconds we are told the velocity is equal to \(\displaystyle k\). 
 \(\displaystyle v(t=10)=k\: \: \: \: \: \:\rightarrow\: \: \: \: \: \:\frac{1}{2}(a+b)100+v_o =k\)

\(\displaystyle 50(a+b)+v_o =k\)

\(\displaystyle v_o =k-50(a+b)\)

The average value of a function on a given interval \(\displaystyle [a,b]\) is given by the following function:

\(\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx.\)

Now, let's simply input our values and function in:

\(\displaystyle \frac{1}{\pi-0}\int_{0}^{\pi}-cosxdx=\frac{1}{\pi}[sin(\pi)-sin(0)]=\frac{0}{\pi}*0=0.\)

The position function describing any object is the antiderivative of the velocity function (in other words, velocity is the derivative of position). 

So, we first integrate the velocity function:

\(\displaystyle s(t)=\int v(t)dt= \int (t^2+t+16)dt=\frac{t^3}{3}+\frac{t^2}{2}+16t+C\)

The following rule was used for integration:

\(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\)

Now, to determine the constant of integration, we use our initial condition given,

\(\displaystyle s(0)=10\)

Plugging this into our function, we get

\(\displaystyle 0+0+0+C=10\)

\(\displaystyle C=10\)

Our final answer is

\(\displaystyle \frac{t^3}{3}+\frac{t^2}{2}+16t+10\)