To find the average value, we must take the integral of f(x) between 3 and 6 and then multiply it by 1/(6 – 3) = 1/3.

The indefinite form of the integral is: 3x⁴ + 7.5x² + 5x

The integral from 3 to 6 is therefore: (3(6)⁴ + 7.5(6)² + 5(6)) - (3(3)⁴ + 7.5(3)² + 5(3)) = (3888 + 270 + 30) – (243 + 22.5 + 15) = 3907.5

The average value is 3907.5/3 = 1302.5

To find the dot product, we multiply the individual corresponding components and add.  

Here, the dot product is found by:

2 * 5 + 2 * (-3) + (-1) * 2 = 2.

The limits of the integration are found by solving  \(\displaystyle y = 2 - x^2\) and \(\displaystyle y=-x\) for \(\displaystyle x\):
\(\displaystyle 2-x^2=-x\)
\(\displaystyle x^2-x-2=0\)
\(\displaystyle (x+1)(x-2)=0\)
\(\displaystyle x=-1, x=2\)

The region runs from \(\displaystyle x=-1\) to \(\displaystyle x=2\). The limits of the integration are \(\displaystyle a=-1\), \(\displaystyle b=2\). 
The area between the curves is:

\(\displaystyle A = \int_{a}^{b} [f(x)-g(x)]dx = \int_{-1}^{2}[(2-x^2)-(-x))]dx\)
\(\displaystyle = \int_{1}^{2}(2+x-x^2)dx = \left [ 2x+x^2/2-x^3/3\right ]_{-1}^{2}\)
\(\displaystyle = (4+\frac{4}{2}-\frac{8}{3}) - (-2 +\frac{1}{2}+\frac{1}{3}) = \frac{9}{2}\)

\(\displaystyle \int \frac{3x^2-2x+1}{\sqrt{x}}dx=\int (3x^{3/2}-2x^{1/2}+x^{-1/2})dx \Rightarrow\)

\(\displaystyle 3(\frac{2}{5})x^{5/2}-2(\frac{2}{3})x^{3/2}+2x^{1/2}+C = \frac{6}{5}x^{5/2}-\frac{4}{3}x^{3/2}+2x^{1/2}+C\)

\(\displaystyle =\frac{2}{5}\sqrt{x}(9x^2-10x+15)+C\)

\(\displaystyle \int \frac{1}{sec(c)}dx=\int cos(x)dx=sin(x)+C\)

\(\displaystyle y = x\) is only above \(\displaystyle y = x^2\) over the interval \(\displaystyle (0,1)\).  Areas are given by the definite integral of each function\(\displaystyle a_1 = \int_{0}^{1}xdx\) and \(\displaystyle a_2 = \int_{0}^{1}x^2dx\)

The area between the curves is found by subtracting the area between each curve and the \(\displaystyle x\)-axis from each other.  For \(\displaystyle y = x\) this area is \(\displaystyle \frac{1}{2}\) and for \(\displaystyle y = x^2\) the area is \(\displaystyle \frac{1}{3}\) giving an area between curves of \(\displaystyle \frac{1}{6}\)

To find the area below a curve, you must find the definite integral of the function.  In this case the limits of integration are where the original function intercepts the \(\displaystyle x\)-axis at \(\displaystyle -2\) and \(\displaystyle 2\).  So you must find \(\displaystyle \int_{-2}^{2}(-x^2+4)dx\) which is \(\displaystyle -\frac{1}{3}x^3+4 x\) evaluated from \(\displaystyle -2\) to \(\displaystyle 2\).  This gives an answer of \(\displaystyle \frac{32}{3}\)

To solve this problem, we first need to find the point where the two equations are equal.  Doing this we find that

 \(\displaystyle 2x=x^2\Rightarrow x^2-2x=0\Rightarrow x(x-2)=0\).  

From this, we see that the two graphs are equal at \(\displaystyle x=0\) and \(\displaystyle x=2\).  We also know that for \(\displaystyle 0< x< 2\), \(\displaystyle y=2x\) is greater than \(\displaystyle y=x^2\).  

So to find the area between these curves we need to evaluate the integral \(\displaystyle \int_{0}^{2} (2x-x^2)dx\).  

The solution to the integral is

\(\displaystyle \int(2x-x^2)dx=x^2-\frac{1}{3}x^3\).

Evaluating this at \(\displaystyle x=2\) and \(\displaystyle x=0\) we get

\(\displaystyle (2*2-\frac{1}{3}2^3)-(0-0)=\frac{12}{3}-\frac{8}{3}=\frac{4}{3}.\)

To solve this problem, we will need to do a \(\displaystyle u\)-substitution.  Letting

 \(\displaystyle u=-x^2\Rightarrow \frac{du}{dx}=-2x\Rightarrow \frac{-du}{2}=xdx\).  

Substituting our \(\displaystyle u(x)\) function back into the integral, we get

 \(\displaystyle \int \frac{-1}{2}e^udu=\frac{-1}{2}e^u=\frac{-1}{2}e^{-x^2}.\)

Evaluating this at \(\displaystyle x=\infty\) and \(\displaystyle x=0\) we get

 \(\displaystyle \frac{-1}{2}e^{- \infty^2}-(\frac{-1}{2}e^0)=0-(\frac{-1}{2})=\frac{1}{2}\)

The average is given by integration as:

\(\displaystyle \text{avg}_{A, B}[ f ] = \frac{1}{B - A} \int_A^B f(x)~dx\)

This means that:

\(\displaystyle \text{avg}_{0, 1}[x \to x^3 - 2x^2 + x ] = \frac{1}{1 - 0} \int_0^1 [x^3 - 2x^2 + x]~dx\)

\(\displaystyle = \left[\frac{x^4}{4} - \frac{2 x^3}{3} + \frac{x^2}{2}\right]_0^1 = \frac{1}{4} - \frac{2}{3} + \frac{1}{2} = \frac{1}{12}\)