In this particular problem, the bounds for our integral are provided; specifically, we know that the integral setup should look something like $\displaystyle \int _0^1(upper function)-(lower function) dx$. This can be determined by the given domain $\displaystyle 0\leq x\leq 1$ and the fact that we are dealing with an area problem involving two functions. 

The next step is to continue creating the integral that will be used to evaluate the area. By graphing the functions, it can be determined that the function $\displaystyle y=x^2+1$ takes on larger values of $\displaystyle y$ than the other function $\displaystyle y=sin(x)$, and therefore should be deemed the “upper function.”

Without graphing, this can also be determined because $\displaystyle y=sin(x)$ will take on values between $\displaystyle 0$ and $\displaystyle 1$, while $\displaystyle x^2+11$. 

The following integral is then created:

$\displaystyle \int_{0}^{1}(x^2+1)-(sin(x)) dx$

Finally, the definite integral is evaluated:

$\displaystyle \int_{0}^{0}(x^2+1)-(sin(x)) dx$$\displaystyle =[\frac{1}{3}x^3+x+cos(x)] |_0^1=(\frac{1}{3}+1+cos(1))-(0+0+1)=\frac{1}{3}+cos(1)=0.87$

First, the intersection points of the two functions must be found in order to determine the bounds of the area integral we wish to set up. 

$\displaystyle y=5x^3$ and $\displaystyle y=3x^2$

$\displaystyle 5x^3=3x^2$

$\displaystyle x={0,0.6}$

Next, the area integral must be set up correctly. To do this, the two functions must be either categorized as an “upper function” or a “lower function.” This is important as it determines which function comes first in our general formula for the area between two curves, $\displaystyle \int_{a}^{b} (upper function)-(lower function) dx$. 

After examination either numerically or graphically, $\displaystyle y=3x^2$ is determined to be above the other function, $\displaystyle y=5x^3$. Now, the integral may be set up:

$\displaystyle \int_{0}^{0.6} (3x^2)-(5x^3) dx$

Finally, the integral must be evaluated:

$\displaystyle Area = \int_{0}^{0.6}(3x^2)-(5x^3) dx = [x^3-\frac{5}{4}x^4] |_0^{0.6} = (0.6)^3-\frac{5}{4}(0.6)^4 = 0.05$

This problem can be tricky to set up. There are two main methods for setting up definite integrals meant to evaluate the area between curves (see below). The setup is often dictated by the potential ease of evaluation.

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

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

Because one of the functions contains a $\displaystyle y^2$ term, the expression may be easier to evaluate if we orient the integral in terms of $\displaystyle dy$. However, note that both methods are valid for this particular problem.

If the method chosen is the integral in terms of $\displaystyle dy$, the next step is determining the bounds. Remember that this process is the same as usual, but instead of bounds in terms of x we would like them in terms of $\displaystyle y$.

$\displaystyle y=x+1 \rightarrow x=y-1$

Now, set the equations equal and solve for intersection points that will determine the bounds.

$\displaystyle y-1=\frac{1}{2}y^2-1$

$\displaystyle y={0,2}$

Finally, set up the integral with the format $\displaystyle \int_{a}^{b}(right function)-(left function) dy$ and solve for the area.

$\displaystyle Area=\int_{0}^{2}(y-1)-(\frac{1}{2}y^2-1) dy$

$\displaystyle Area= [\frac{1}{2}y^2-y-\frac{1}{6}y^3+y] |_0^2 = [\frac{1}{2}y^2-\frac{1}{6}y^3] |_0^2 = 2-\frac{8}{6}-0+0 = \frac{2}{3}$

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$

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