Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

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

Solution:

First, simplify the function and then evaluate the integral.

1. Simplify

\(\displaystyle (x-4)^2+4\)

\(\displaystyle x^2+16-8x+4\)

\(\displaystyle x^2-8x+20\)

2. Evaluate the integral

\(\displaystyle \int_{-1}^{4} x^2-8x+20dx\)

\(\displaystyle \frac{1}{3}x^3-4x^2+20x\Big|_{-1}^{4}\)

\(\displaystyle \frac{1}{3}\cdot64-4\cdot16+80-(-\frac{1}{3}-4-20)\)

\(\displaystyle \frac{64}{3}-64+80+\frac{1}{3}+24\)

\(\displaystyle \frac{65}{3}+40 = \frac{185}{3} = 61.667\)

When rounded tot he nearest integer, the area under the curve is \(\displaystyle 62\)

We can represent the area as:

\(\displaystyle \int_{0}^{2\pi}tan(x) dx= \int_{0}^{2\pi}\frac{sin(x)}{cos(x)} dx\)

\(\displaystyle u=cos(x)\), \(\displaystyle \frac{du}{dx}=-sin(x)\)

\(\displaystyle \int \frac{sin(x)}{cos(x)} dx= \int -\frac{1}{u}du=-ln(u)+C\)

\(\displaystyle F(x)= -ln(cos(x)) +C\)

By the fundamental theorem of calculus:

\(\displaystyle \int_{0}^{2\pi} \frac{sin(x)}{cos(x)} dx= F(2\pi)-F(0)= -ln(cos(2\pi))+ln(cos(0))\)

                              \(\displaystyle =0+0=0\)

Looking at the inside integral: 

\(\displaystyle \int_{0}^{5}1 dx= 5-0=5\)

Having done the inside integral, we can do the outside integral

\(\displaystyle \int_{0}^{5}5 dy= 5*5-5*0=25\)

The area between two curves \(\displaystyle y=f_1(x)\) and \(\displaystyle y=f_2(x)\) is given by the formula \(\displaystyle A=\int^b_a(f_1(x)-f_2(x))dx\), where \(\displaystyle f_1(x)\) is the upper bound curve, \(\displaystyle f_2(x)\) is the lower bound curve, and \(\displaystyle a\) and \(\displaystyle b\) are the solutions to the equation \(\displaystyle f_1(x)=f_2(x)\). The graphs of \(\displaystyle f(x)=x^3\) and \(\displaystyle g(x)=\sqrt[3]{x}\) are given in the figure below:

As we can see, the graph of \(\displaystyle g(x)\) is the upper bound for the area and the graph of \(\displaystyle f(x)\) is the lower bound for the area. It is also apparent that the limits of integration are given by \(\displaystyle a=0\) and \(\displaystyle b=1\). To see this algebraically, for graphs often do not give us clear answers for limits of integration, we would solve the equation \(\displaystyle f(x)=g(x)\). Plugging in \(\displaystyle f(x)\) and \(\displaystyle g(x)\), we obtain:

\(\displaystyle x^3=\sqrt[3]{x}\)

\(\displaystyle x^3=x^{1/3}\)

\(\displaystyle x^3-x^{1/3}=0\)

\(\displaystyle x^{1/3}(x^{2/3}-1)=0\)

Setting each factor equal to \(\displaystyle 0\) gives us \(\displaystyle x^{1/3}=0\rightarrow x=0\), and \(\displaystyle x^{2/3}-1=0\rightarrow x^{2/3}=1\rightarrow x=1\). With these limits of integration, our integral for the area becomes:

\(\displaystyle A=\int_0^1(x^{1/3}-x^3)dx=\frac{3}{4}x^{4/3}-\frac{x^4}{4}\biggr|_0^1\)

\(\displaystyle =\biggr(\frac{3}{4}(1^{4/3})-\frac{1}{4}\biggr(\frac{1^4}{4}\biggr)\biggr)-\biggr(\frac{3}{4}(0^{4/3})-\frac{1}{4}\biggr(\frac{0^4}{4}\biggr)\biggr)\)

\(\displaystyle =\biggr(\frac{3}{4}-\frac{1}{4}\biggr)-0=\frac{1}{2}\)

Therefore the area between the curves must be \(\displaystyle \frac{1}{2}\).

To calculate the area between two functions, take the integral of the function on top minus the function on bottom. The intersection of the functions will be the bound on the integral. In this particular case the top function is the parabola described as \(\displaystyle 9-x^2\) and the bottom function is the x-axis, or in other words zero.

Therefore the basic integral looks as follows.

\(\displaystyle \int (9-x^2)-0dx\)

The question indicates the upper and lower bound as \(\displaystyle x=3\) and \(\displaystyle x=0\), applying these bounds to the integral results in,

\(\displaystyle \int_{0}^{3}(9-x^2)dx\).

To calculate the integral recall the following rule of integration.

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

where C represents a constant.

Applying this rule to the integral in question results in,

\(\displaystyle (9x-\frac{x^3}{3})|_{0}^{3}\)

Substituting in the bounds results in the solution for area.

\(\displaystyle 9(3)-\frac{3^3}{3}=18\)

Find the area bounded by the functions: 

\(\displaystyle g(x)=x+6\)

\(\displaystyle f(x)=2(x+3)^2-1\)

Looking at the plot of the function we can see that \(\displaystyle g(x)\geq f(x)\) for all \(\displaystyle x\) in the region. The area formula is therefore, 

\(\displaystyle A=\int_{a}^{b}|g(x)-f(x)|dx\)

, for \(\displaystyle g(a)=f(a)\) and \(\displaystyle g(b)=f(b)\)

To find the intersection points, set the functions equal to each other and solve for \(\displaystyle x.\)

\(\displaystyle g(x)=f(x)\)

\(\displaystyle x+6=2(x+3)^2-1\)

\(\displaystyle x+6=2(x^2+6x+9)-1\)

\(\displaystyle x+6=2x^2+12x+17\)

\(\displaystyle 2x^2+11x+11=0\)

\(\displaystyle x =\frac{-11\pm\sqrt{11^2-4(11)(2))}}{2(2)}\)

\(\displaystyle =-\frac{11}{4}\pm \frac{\sqrt{33}}{4}\)

Therefore, the two lines intersect for the following values of \(\displaystyle x\): 

\(\displaystyle x = \left \{ -\frac{11}{4}-\frac{\sqrt{33}}{4},- \frac{11}{4}+\frac{\sqrt{33}}{4}\right \}\).

So we will integrate over this interval, 

 \(\displaystyle A=\int_{-\frac{11}{4}-\frac{\sqrt{33}}{4}}^{-\frac{11}{4}+\frac{\sqrt{33}}{4}}|(x+6)-[2(x+3)^2-1])dx\)

\(\displaystyle =\int_{-\frac{11}{4}-\frac{\sqrt{33}}{4}}^{-\frac{11}{4}+\frac{\sqrt{33}}{4}}[(x+6)-(2x^2+12x+17)]dx\)

Finally we simplify the integrand to arrive at: 

\(\displaystyle =\int_{-\frac{11}{4}-\frac{\sqrt{33}}{4}}^{-\frac{11}{4}+\frac{\sqrt{33}}{4}}(-2x^2-11x-11)dx\)

If you were to evaluate the integration is simple, but evaluating the limits would be quite tedious. The outcome is: 

\(\displaystyle =\left [ -\frac{2}{3}x^3-\frac{11}{2}x^2-11x\right ]_{\frac{11}{4}-\frac{\sqrt{33}}{4}}^{\frac{11}{4}+\frac{\sqrt{33}}{4}}\)

\(\displaystyle =\frac{11\sqrt{33}}{8}\)

When setting up this problem, it should look like this: \(\displaystyle \int_{2}^{4}x^2-1dx\). Then, integrate. Remember that when integrating, raise the exponent by 1 and then also put that result on the denominator: \(\displaystyle \frac{x^3}{3}-x\). Then, evaluate first at 4 and then at 2. Subtract the results. \(\displaystyle (\frac{64}{3}-4)-(\frac{8}{3}-2)=\frac{56}{3}-2=\frac{50}{3}\).

First, set up the integral expression: \(\displaystyle \int_{4}^{5}x^2-3x-4dx\). Then, integrate. Remember, when integrating, raise the exponent by 1 and then put that result on the denominator: \(\displaystyle \frac{x^3}{3}-\frac{3x^2}{2}-4x\). Then, evaluate at 5 and then 4. Subtract those results: \(\displaystyle (\frac{125}{3}-\frac{75}{2}-20)-(\frac{64}{3}-\frac{48}{2}-16)\). Simplify to get your final answer of \(\displaystyle \frac{17}{6}\).

First, set up the integral expression: \(\displaystyle \int_{2}^{3}x^3-2x+1dx\). Then, integrate. Remember to raise the exponent by 1 and then put that result on the denominator: \(\displaystyle \frac{x^4}{4}-\frac{2x^2}{2}+x\). Evaluate at 3 and then 2. Subtract those 2 results: \(\displaystyle (\frac{81}{4}-9+3)-(\frac{16}{4}-4+2)=\frac{65}{4}-4=\frac{49}{4}\).

First, set up the integral expression:
\(\displaystyle \int_{4}^{5}x^2-3x+1dx\)

Now, integrate:

\(\displaystyle \frac{x^3}{3}-\frac{3x^2}{2}+x\)

Evaluate at 5 and 4. Subtract the results:

\(\displaystyle (\frac{125}{3}-\frac{75}{2}+5)-(\frac{64}{3}-\frac{48}{2}+4)=(\frac{125}{3}-\frac{75}{2}+5)\)

Simplify to get:

\(\displaystyle \frac{47}{6}\)