For this particular function we will need to preform a "u-substitution".

In our case let

\(\displaystyle \\u=ln(x) \\\) which will make \(\displaystyle du=\frac{1}{x}\).

Now we will substitute these into our integral to get the following.

\(\displaystyle \int_{1}^{e}\frac{2ln(x)}{x}=\int_{1}^{e} 2U du\) if  \(\displaystyle U=ln(x)\) 

Then we integrate using the power rule which states,

\(\displaystyle x^n\rightarrow nx^{n-1}\)

\(\displaystyle \int_{1}^{e}2Udu= U^2\)

Now plug back in the original variable and then subtract the function values at the bounds.

\(\displaystyle = ln^2(e)-ln^2(1)=1-0=1\)

To find the area under the curve, we must integrate over the given bounds:

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

To integrate, we must do the following substitution:

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

The derivative was found using the following rule:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\)

Now, rewrite the integral, and change the bounds in terms of u:

\(\displaystyle \int_{-1}^{1}u^2du\)

Note that during the rewriting, the bounds changed to the upper bound being -1 and the lower bound being 1, but the negative sign that came from the derivative of u allowed us to make the bounds as they are seen above. 

Now perform the definite integration:

\(\displaystyle \int_{-1}^{1}u^2du=\frac{1}{3}-\left(-\frac{1}{3}\right)=\frac{2}{3}\)

The integral was found using the following rule:

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

and the definite integration was finished by plugging in the upper bound into the resulting function, plugging the lower bound into the resulting function, and subtracting the two, as seen above. 

To find the area underneath the curve of the function \(\displaystyle f(x)\) on the interval \(\displaystyle [a,b]\), we find the definite integral

\(\displaystyle \int_{a}^{b}|f(x)|\, dx\)

Because the function in this problem is always positive on the interval, or

\(\displaystyle f(x)>0\) on the interval \(\displaystyle [0,9]\) the area underneath the curve can be found using the definite integral

\(\displaystyle \int_{0}^{9}\sqrt{x}\,dx\)

and rewriting the function the definite integral is

\(\displaystyle \int_{0}^{9}x^{\frac{1}{2}}\,dx\)

Using the inverse power rule which states

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

we find the definite integral to be

\(\displaystyle \int_{0}^{9}x^{\frac{1}{2}}\,dx=\frac{2}{3}x^{\frac{3}{2}}|_{0}^{9}\)

And by the corollary of the Fundamental Theorem of Calculus, the definite integral equals

\(\displaystyle \frac{2}{3}(9)^{\frac{3}{2}}-\frac{2}{3}(0)^{\frac{3}{2}}=18\)

As such, the area is \(\displaystyle 18\) square units

We can define the area underneath a curve provided by a function as the definite integral of the function over a given space. Thus, given \(\displaystyle y=3x^{2}-1\), then the area over \(\displaystyle 0\leq x\leq 3\) is \(\displaystyle \int_{0}^{3}(3x^{2}-1)dx\). 

Using the Power Rule for Integrals

for all ,

we can determine that:

\(\displaystyle \int_{0}^{3}(3x^{2}-1)dx\)

 \(\displaystyle =[x^{3}-x]_{0}^{3}\textrm{}\)

\(\displaystyle =[(3)^{3}-3]-[0^{3}-0]\)

\(\displaystyle =27-3\)

\(\displaystyle =24\)

We can define the area underneath a curve provided by a function as the definite integral of the function over a given space. Thus, given \(\displaystyle y=\frac{1}{2}x^{2}+x\), then the area over \(\displaystyle 0\leq x\leq 2\) is \(\displaystyle \int_{0}^{2}(\frac{1}{2}x^{2}+x)dx\). 

Using the Power Rule for Integrals

 for all ,

we can determine that:

\(\displaystyle \int_{0}^{2}(\frac{1}{2}x^{2}+x)dx\)

 \(\displaystyle =[\frac{1}{6}x^{3}+\frac{1}{2}x^{2}]_{0}^{2}\textrm{}\)

\(\displaystyle =[\frac{1}{6}(2)^{3}+\frac{1}{2}(2)^{2}]-[\frac{1}{6}(0)^{3}+\frac{1}{2}(0)^{2}]\)

\(\displaystyle =[\frac{8}{6}+2]\)

\(\displaystyle =[\frac{8}{6}+\frac{12}{6}]\)

\(\displaystyle =\frac{20}{6}\)

\(\displaystyle =\frac{10}{3}\)

We can define the area underneath a curve provided by a function as the definite integral of the function over a given space. Thus, given \(\displaystyle y=6x^{2}-2x\), then the area over \(\displaystyle 0\leq x\leq 4\) is \(\displaystyle \int_{0}^{4}(6x^{2}-2x)dx\). 

Using the Power Rule for Integrals

 for all ,

we can determine that:

\(\displaystyle \int_{0}^{4}(6x^{2}-2x)dx\)

\(\displaystyle =[2x^{3}-x^{2}]_{0}^{4}\textrm{}\)

\(\displaystyle =[2(4)^{3}-(4)^{2}] -[2(0)^{3}-(0)^{2}]\)

\(\displaystyle =[2(64)-16]\)

\(\displaystyle =128-16\)

\(\displaystyle =112\)

We can define the area underneath a curve provided by a function as the definite integral of the function over a given space.

Thus, given \(\displaystyle y=\frac{1}{2}x^{2}-3x\), then the area over \(\displaystyle 0\leq x\leq 2\) is \(\displaystyle \int_{0}^{2}\left(\frac{1}{2}x^{2}-3x\right)dx\).

Using the Power Rule for Integrals

for all ,

we can determine that:

\(\displaystyle \int_{0}^{2}\left(\frac{1}{2}x^{2}-3x\right)dx\)

\(\displaystyle =\left[\frac{1}{6}x^{3}-\frac{3}{2}x^{2}\right]_{0}^{2}\textrm{}\)

\(\displaystyle =\left[\frac{1}{6}(2)^{3}-\frac{3}{2}(2)^{2}\right]-\left[\frac{1}{6}(0)^{3}-\frac{3}{2}(0)^{2}\right]\)

\(\displaystyle =\left[\frac{8}{6}-\frac{12}{2}\right]\)

\(\displaystyle =\left[\frac{8}{6}-6\right]\)

\(\displaystyle =\frac{8}{6}-\frac{36}{6}\)

\(\displaystyle =-\frac{28}{6}=-\frac{14}{3}\)

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:

\(\displaystyle \int_{-\pi/2}^{\pi/2}2-cos(x)dx\)

Solution:

\(\displaystyle \int2-cos(x)dx = 2x-sin(x)\)

\(\displaystyle 2\pi/2 - sin(\pi/2)-(-2\pi/2-sin(-\pi/2) =\)

\(\displaystyle 2\pi-2\)

Note:

\(\displaystyle sin(\pi/2) = 1\)

\(\displaystyle \int cos(x) = sin(x)\)

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:

\(\displaystyle \int_{3}^{6}3x+2dx\)

Solution:

\(\displaystyle \int_{3}^{6} 3x+2dx\)

\(\displaystyle =\frac{3}{2}x^2+2x\Big|_{3}^{6}\)

\(\displaystyle =\frac{3}{2}\cdot 6^2+2\cdot6 -(\frac{3}{2}\cdot 3^2+2\cdot 3)\)

\(\displaystyle =\frac{3}{2}\cdot36+12-\frac{3}{2}\cdot9-6\)

\(\displaystyle =\frac{93}{2} = 46.5\) 

\(\displaystyle 47\) after rounding

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:

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

Solution:

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

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

\(\displaystyle =\frac{2}{3}\cdot 4^3+4\cdot4 -(\frac{2}{3}1^3+4)\)

\(\displaystyle =\frac{2}{3}\cdot 64+16-\frac{2}{3}-4=54\)