# Thread: Integral, log and Inverse

1. ## Integral, log and Inverse

1.$\displaystyle \int_1^2 \frac{x-4}{x^2} dx$ *ANSWERED*

2.$\displaystyle log_a(2^a) = \frac{a}{4}$, then $\displaystyle a =$ * ANSWERED*

3.Inverse of $\displaystyle y=2x^3+3x^2+7x+4$ *ANSWERED*
(i know how to do inverse, but this polynomial is throwing me off...)

2. Originally Posted by yamuda
1.$\displaystyle \int_0^2 \frac{x-4}{x^2} dx$

2.$\displaystyle log_a(2^a) = \frac{a}{4}$, then $\displaystyle a =$

3.Inverse of $\displaystyle y=2x^3+3x^2+7x+4$
(i know how to do inverse, but this polynomial is throwing me off...)

Here is number 2

rewrite the left hand side using the change of base formula

$\displaystyle \log_a2^a=\frac{\log_22^a}{\log_2a}$

Now we get

$\displaystyle \frac{\log_22^a}{\log_2a}=\frac{a}{4} \iff \frac{a}{\log_2a}=\frac{a}{4}$

$\displaystyle 4=\log_2a \iff 2^4=a$

Good luck.

3. Originally Posted by yamuda
1.$\displaystyle \int_0^2 \frac{x-4}{x^2} dx$ *NO ANSWER YET*

2.$\displaystyle log_a(2^a) = \frac{a}{4}$, then $\displaystyle a =$ * ANSWERED*

3.Inverse of $\displaystyle y=2x^3+3x^2+7x+4$ *NO ANSWER YET*
(i know how to do inverse, but this polynomial is throwing me off...)

for the first use termwise division

$\displaystyle \int_{0}^{2}\frac{x-4}{x^2}dx=\lim_{a \to 0} \left(\int_{a}^{2}\frac{1}{x}dx-\int_a^2\frac{4}{x^2}dx \right)$

after integrating and evaluating the limit you will find that it divereges to infinity.

4. ## hmmm...

For the last one are supposed to find the inverse of just show that an inverses exisis?

Here is a link to the general cubic formula( like the quadratic, but for degree 3 equations)

The Cubic Formula

Good luck

5. Originally Posted by TheEmptySet
for the first use termwise division

$\displaystyle \int_{0}^{2}\frac{x-4}{x^2}dx=\lim_{a \to 0} \left(\int_{a}^{2}\frac{1}{x}dx-\int_a^2\frac{4}{x^2}dx \right)$

after integrating and evaluating the limit you will find that it divereges to infinity.
mmm... i messed up on the integral... was supposed to be $\displaystyle \int_1^2$, im tryna figure out the integral, coming out with a numerical answer, not if it goes to infinity..

and for the last one, im trying to find the inverse, so i switched y and x, then i have to solve for y
this is all i got... $\displaystyle x=2y^3+3y^2+7y+4$

6. Originally Posted by yamuda
mmm... i messed up on the integral... was supposed to be $\displaystyle \int_1^2$, im tryna figure out the integral, coming out with a numerical answer, not if it goes to infinity..

and for the last one, im trying to find the inverse, so i switched y and x, then i have to solve for y
this is all i got... $\displaystyle x=2y^3+3y^2+7y+4$
$\displaystyle \int_{1}^{2}\frac{x-4}{x^2}dx=\int_{1}^{2}\frac{1}{x}dx-\int_1^2\frac{4}{x^2}dx=\ln|x|+\frac{4}{x}|_1^2$

$\displaystyle \ln2+\frac{4}{2}- \left( \ln1+\frac{4}{1}\right)=\ln2-2$

as for your last question the only way I know of is the formula from the above link.

Good luck

7. Originally Posted by TheEmptySet
$\displaystyle \int_{1}^{2}\frac{x-4}{x^2}dx=\int_{1}^{2}\frac{1}{x}dx-\int_1^2\frac{4}{x^2}dx=\ln|x|+\frac{4}{x}|_1^2$

$\displaystyle \ln2+\frac{4}{2}- \left( \ln1+\frac{4}{1}\right)=\ln2-2$

as for your last question the only way I know of is the formula from the above link.

Good luck
alright, thanks for all the help!!!!