1. ## simple definite integral

$\displaystyle \int_2^{6} \frac {6(x^2+1)}{ \sqrt{x}} dx$

I'm having a bit of trouble. I know I can draw out the 6 and bring the denominator up, but how do I follow that up?

2. $\displaystyle \frac{{6\left( {x^2 + 1} \right)}} {{\sqrt x }} = 6\left( {x^{\frac{3} {2}} + x^{ - \frac{1} {2}} } \right)$

3. Make substitution

$\displaystyle x=z^2$

$\displaystyle dx=2zdz$

4. I think you might be confusing the issue a little by substituting z for x, as x is the original variable in the function. Plato's way makes it easier, as you can pull the 6 out in front of the integral, then evaluate using the exponent rule for integrals.

5. Originally Posted by Plato
$\displaystyle \frac{{6\left( {x^2 + 1} \right)}} {{\sqrt x }} = 6\left( {x^{\frac{3} {2}} + x^{ - \frac{1} {2}} } \right)$
Not sure how you got this. The x^-(1/2) is from the denominator, yeah, but what's that first term? What happened to the x^2+1?

6. $\displaystyle \frac{{6\left( {x^2 + 1} \right)}}{{\sqrt x }} = 6(x^2 + 1) \frac{1}{x^{\frac{1}{2}}} = 6\left( {(x^{4 \over 2} + 1) x^{ {-}{1\over 2}} } \right) = 6\left( {x^{\frac{3}{2}} + x^{ - \frac{1}{2}} } \right)$

Makes sense?

7. Originally Posted by Archduke01
Not sure how you got this. The x^-(1/2) is from the denominator, yeah, but what's that first term? What happened to the x^2+1?
Very basic “calculus confusion” is almost always the result of someone who is ‘algebra-deficient’.

8. Originally Posted by Archduke01
Not sure how you got this. The x^-(1/2) is from the denominator, yeah, but what's that first term? What happened to the x^2+1?
When you divide a variable with an exponent by the same variable with a different exponent you subtract the exponent of the denomintor from the numerator.

So

$\displaystyle x^2/x$ = x^(2-1) = x

These math tags are killing me...how do i properly show that an exponent is ^(2-1), it doesnt display like this :P

9. Originally Posted by AllanCuz
$\displaystyle x^2/x$ = x^(2-1) = x
These math tags are killing me...how do i properly show that an exponent is ^(2-1), it doesnt display like this :P
$$x^{2-1}$$ gives $\displaystyle x^{2-1}$.
Put the exponent in {}

10. Originally Posted by losm1
$\displaystyle \frac{{6\left( {x^2 + 1} \right)}}{{\sqrt x }} = 6(x^2 + 1) \frac{1}{x^{\frac{1}{2}}} = 6\left( {(x^{4 \over 2} + 1) x^{ {-}{1\over 2}} } \right) = 6\left( {x^{\frac{3}{2}} + x^{ - \frac{1}{2}} } \right)$

Makes sense?
EDIT: Nevermind I see my mistake.