First, divide up into three different expressions so you can integrate each x term separately:

\(\displaystyle \int2x^{4}dx-\int6x^{3}dx+\int9xdx\)

Then, integrate and simplify:

\(\displaystyle 2(\frac{x^{5}}{5})-6(\frac{x^4}{4})+9(\frac{x^2}{2})\)

\(\displaystyle \frac{2x^5}{5}-\frac{3x^4}{2}+\frac{9x^2}{2}\)

Don't forget "C" because it's an indefinite integral: 

\(\displaystyle \frac{2x^5}{5}-\frac{3x^4}{2}+\frac{9x^2}{2}+C\)

To start the problem, it's easier if you bring up the denominator and make it a negative exponent:

\(\displaystyle \int x^{-3}dx\)

Then, integrate:

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

Simplify and add the "C" for an indefinite integral:

\(\displaystyle -\frac{1}{2x^2}+ C\)

Plug in the initial conditions [F(1)=0] to find C and generate the particular solution:

\(\displaystyle -\frac{1}{2x^2}+C=0\)

\(\displaystyle -\frac{1}{2(1^2)}+C=0\)

\(\displaystyle C=\frac{1}{2}\)

Thus, your final equation is: 

\(\displaystyle F(x)=-\frac{1}{2x^2}+\frac{1}{2}\)

First, split up into 2 integrals:

\(\displaystyle \int2xdx-\int3x^2dx\)

Then integrate and simplify:

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

\(\displaystyle x^2-x^3\)

Don't forget to add C because it's an indefinite integral:

\(\displaystyle -x^3+x^2 +C\)

First, FOIL the binomial:

\(\displaystyle (x^3+2)(x^3+2)= x^6+4x^3+4\)

Once that's expanded, integrate each piece separately:

\(\displaystyle \int x^6dx+\int4x^3dx+\int4dx\)

\(\displaystyle \frac{x^7}{7}+4(\frac{x^4}{4})+4x\)

Then simplify and add C because it's an indefinite integral:

\(\displaystyle \frac{x^7}{7}+x^4+4x+C\)

Remember the Rundamental Theorem of Calculus: If \(\displaystyle g'(x)=f(x)\), then \(\displaystyle \int^b_af(x){\mathrm{d} x}=g(b)-g(a)\).

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

\(\displaystyle \int x^2+5x{\mathrm{d} x}=\frac{x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}+c\)

Remember, when taking an integral, definite or indefinite, we always add \(\displaystyle +c\), as there could be a constant involved.

\(\displaystyle \int x^2+5x{\mathrm{d} x}=\frac{1}{3}x^3+\frac{5}{2}x^2+c\)

Now we can plug that back into the problem.

\(\displaystyle \int^b_a x^2+5x{\mathrm{d} x}=(\frac{1}{3}b^3+\frac{5}{2}b^2+c)-(\frac{1}{3}a^3+\frac{5}{2}a^2+c)\)

Notice that the \(\displaystyle c\)'s cancel out. Plug in the values given in the problem:

\(\displaystyle \int^6_3 x^2+5x{\mathrm{d} x}=(\frac{1}{3}(6)^3+\frac{5}{2}(6)^2)-(\frac{1}{3}(3)^3+\frac{5}{2}(3)^2)\)\(\displaystyle \int^6_3 x^2+5x{\mathrm{d} x}=(\frac{1}{3}*216+\frac{5}{2}*36)-(\frac{1}{3}*27+\frac{5}{2}*9)\)

\(\displaystyle \int^6_3 x^2+5x{\mathrm{d} x}=(72+90)-(9+22.5)\)

\(\displaystyle \int^6_3 x^2+5x{\mathrm{d} x}=(162)-(31.5)\)

\(\displaystyle \int^6_3 x^2+5x{\mathrm{d} x}=130.5\)

We can use the substitution technique to evaluate this integral.

Let \(\displaystyle u=1-t\).

We will differentiate \(\displaystyle u\) with respect to \(\displaystyle t\).

\(\displaystyle \frac{\mathrm{d}u }{\mathrm{d} t}=-1\), which means that \(\displaystyle {\mathrm{d} u}=-{\mathrm{d} t}\).

We can solve for \(\displaystyle {\mathrm{d}t }\) in terms of \(\displaystyle {\mathrm{d} u}\), which gives us \(\displaystyle {\mathrm{d}t }=-{\mathrm{d} u}\).

We will also need to change the bounds of the integral. When \(\displaystyle t=-1\), \(\displaystyle u=1-(-1)\), and when \(\displaystyle t=0\), \(\displaystyle u=1-0=1\).

We will now substitute \(\displaystyle u\) in for the \(\displaystyle 1-t\), and we will substitute \(\displaystyle -{\mathrm{d}u }\) for \(\displaystyle {\mathrm{d} t}\).

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

\(\displaystyle \int_{2}^{1}-e^{u}du = -e^{u}|_{2}^{1}=-e^{1}-(-e^{2})=e^{2}-e^{1}\)

The answer is \(\displaystyle e^{2}-e\).

Set \(\displaystyle u = \cos x\).

Then \(\displaystyle \frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \cos x= -\sin x\) and \(\displaystyle \sin x \; dx = - du\).

Also, since \(\displaystyle u = \cos x\), the limits of integration change to \(\displaystyle \cos \frac{\pi }{2} = 0\) and \(\displaystyle \cos 0 = 1\).

Substitute:

\(\displaystyle \dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }\; \; dx\)

\(\displaystyle = \int_{1}^{0} \left (- \sqrt{u} \right ) \; du\)

\(\displaystyle = \int_{1}^{0} \left (-u ^{\frac{1}{2}} \right ) \; du\)

\(\displaystyle = \int_{0}^{1} u ^{\frac{1}{2}} \; du\)

\(\displaystyle \dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \\ & \end{matrix}\right|\) \(\displaystyle \begin{matrix} 1\\ \\ 0 \end{matrix}\)

\(\displaystyle \dpi{150} = \left.\begin{matrix} \\ \frac{ 2u \sqrt{u}} {3} & \\ & \end{matrix}\right|\) \(\displaystyle \begin{matrix} 1\\ \\ 0 \end{matrix}\)

\(\displaystyle =\frac{ 2\cdot 1\cdot \sqrt{1}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} = \frac{ 2} {3}\)

First you must know that:

\(\displaystyle tan^{-1}(x)= \frac{1}{x^2+1}\)  and  \(\displaystyle \int \frac{1}{a^2+x^2}=\bigg(\frac{1}{a}\bigg)tan^{-1}\bigg(\frac{x}{a}\bigg)+C\)

Therefore we can rewrite our problem in this form:

\(\displaystyle \int\frac{1}{x^2+6^2}dx\)

where \(\displaystyle a=6\).

Thus the integral becomes,

\(\displaystyle \int \frac{1}{6^2+x^2}=\bigg(\frac{1}{a}\bigg)tan^{-1}\bigg(\frac{x}{a}\bigg)+C=\bigg(\frac{1}{6}\bigg)tan^{-1}\bigg(\frac{x}{6}\bigg)+C\)

\(\displaystyle \int2x+3 = x^2+3x\)

Setting the limits from zero to two we can find that,

\(\displaystyle \int2x+3 = x^2+3x\)

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

\(\displaystyle =4+6-0\)

\(\displaystyle =10\)

Seeing that the equation contains an absolute value you should know that the graph must always remain positive therefore resulting in a V-shaped graph.

Since the equation is \(\displaystyle y=x-1\), when \(\displaystyle x=1, y=0\) then the vertex of the graph is at \(\displaystyle (1,0)\).

The graph contains a triangle ranging from 0 to 1 and a triangle from 1 to 3. Remebering that taking the interal of a function is the same as finding the area under the curve we can use these triangles to solve our problem.

The area of the triangle from 0 to 1 is,

\(\displaystyle \frac{1}{2}bh=\frac{1}{2}(1)(1)=0.5\).

The area of the triangle from 1 to 3 is,

\(\displaystyle \frac{1}{2}bh=\frac{1}{2}(2)(2)=\frac{4}{2}=2\).

Thus the evaluated integral must be these areas added together,

\(\displaystyle 0.5+2=2.5\).