To find the definite integral, we can use the Fundamental Theorem of Calculus which states that if $\displaystyle g'(x)=f(x)$, then $\displaystyle \int_a^bf(x)=g(b)-g(a)$.

Therefore, we need to find the indefinite integral of our equation to start.

To find the indefinite integral, we can use the reverse power rule. We raise the exponent of the variable by one and divide by our new exponent.

$\displaystyle \int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4x^{2+1}}{2+1}+\frac{\frac{2}{3}x^{1+1}}{1+1}+c$

Remember to include a $\displaystyle +c$ to cover any potential constant that might be in our new equation.

$\displaystyle \int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4x^{3}}{3}+\frac{\frac{2}{3}x^{2}}{2}+c$

$\displaystyle \int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4}{3}x^3+\frac{1}{3}x^2+c$

Plug that into FTOC:

$\displaystyle \int_a^b 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}b^3+\frac{1}{3}b^2+c)-(\frac{4}{3}a^3+\frac{1}{3}a^2+c)$

Notice that the $\displaystyle c$'s cancel out.

Plug in our given values.

$\displaystyle \int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(6)^3+\frac{1}{3}(6)^2)-(\frac{4}{3}(3)^3+\frac{1}{3}(3)^2)$

$\displaystyle \int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(216)+\frac{1}{3}(36))-(\frac{4}{3}(27)+\frac{1}{3}(9))$

$\displaystyle \int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(288+12)-(36+3)$

$\displaystyle \int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(300)-(39)$

$\displaystyle \int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=261$

The fundamental theorem of calculus states that if $\displaystyle f'(x)=g(x)$, then $\displaystyle \int_a^bg(x){\mathrm{d} x}=f(b)-f(a)$.

First, we need to find the indefinite integral of our given equation. Just like with the derivatives, the indefinite integrals or anti-derivatives of trig functions must be memorized.

To solve for the definite integral, use the fundamental theorem of calculus. If $\displaystyle g'(x)=f(x)$, then $\displaystyle \int_a^bf(x)=g(b)-g(a)$.

First we need to find the indefinite integral.

First, bring up the radical into the numerator and distribute to the (x+1) term.

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

$\displaystyle \int x^{\frac{1}{2}} + x^{\frac{-1}{2}}dx$

Then integrate.

$\displaystyle \frac{2}{3}x^{\frac{3}{2}}+ 2x^{\frac{1}{2}}$

Since it's indefinite, don't forget to add the C: 

$\displaystyle \frac{2}{3}x^{\frac{3}{2}}+ 2x^{\frac{1}{2}}+C$

First, divide up into two different integral expressions:

$\displaystyle \int xdx + \int6dx$

Then, integrate each: 

$\displaystyle \frac{x^2}{2} + 6x$

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

$\displaystyle \frac{x^2}{2} + 6x+C$

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)$

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

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

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