# Thread: Two integrals to evaluate

1. ## Two integrals to evaluate

1. $\displaystyle \int_1^e\frac{5x^2+x-3}{x}dx$

For this question, I was wondering if dividing the x through all of the terms would be the best way to start solving, which gives:

$\displaystyle = \int_1^e(5x+1-3x^{-1})dx$ and then taking the antiderivative, evaluating from 1 to e etc...

2. $\displaystyle \int_0^{\frac{\pi}{4}}\frac{d}{dx}(\frac{xtanx}{1+ x})dx$
Also need help starting out this one. For this question does the derivative of $\displaystyle \frac{xtanx}{1+x}$ need to first be taken? I do not know if it would be permissible to pull the d/dx out into the front of the integral, or even if that would help at all.

2. Originally Posted by Em Yeu Anh
1. $\displaystyle \int_1^e\frac{5x^2+x-3}{x}dx$

For this question, I was wondering if dividing the x through all of the terms would be the best way to start solving, which gives:

$\displaystyle = \int_1^e(5x+1-3x^{-1})dx$ and then taking the antiderivative, evaluating from 1 to e etc... Mr F says: Correct.

2. $\displaystyle \int_0^{\frac{\pi}{4}}\frac{d}{dx}(\frac{xtanx}{1+ x})dx$
Also need help starting out this one. For this question does the derivative of $\displaystyle \frac{xtanx}{1+x}$ need to first be taken? I do not know if it would be permissible to pull the d/dx out into the front of the integral, or even if that would help at all.

2. The function you're integrating satisfies the conditions that would allow you to change the order of operations. The answer will be $\displaystyle \left[ \frac{x \tan x}{1 + x}\right]_0^{\pi/4}$.

3. For #2 - don't bother. Just evaluate F(b)-F(a) for your F(x). Good rule of thumb - if it looks ridiculous and crazy, there's no doubt some trick to make it not. I mean you'd have some seriously funky stuff going on here if they honestly expected you to take the integral of that function.

4. Thanks for the help,
I was ill during the lectures that introduced integration, I'm having a lot of difficulties catching up.
Just to clarify the theory behind this, for a continuous function:
$\displaystyle \frac{d}{dx}\int_a^bf(t)dt = f(b) - f(a)$ is that what is happening in that question?

And if either the upper or the lower integration limits involved something in terms of 'x' how would that change the solution?

5. Originally Posted by Em Yeu Anh
[snip]
2. $\displaystyle \int_0^{\frac{\pi}{4}}\frac{d}{dx}(\frac{xtanx}{1+ x})dx$

Also need help starting out this one. For this question does the derivative of $\displaystyle \frac{xtanx}{1+x}$ need to first be taken? I do not know if it would be permissible to pull the d/dx out into the front of the integral, or even if that would help at all.

Originally Posted by mr fantastic
2. The function you're integrating satisfies the conditions that would allow you to change the order of operations. The answer will be $\displaystyle \left[ \frac{x \tan x}{1 + x}\right]_0^{\pi/4}$.
I have made a technical error here that has put you on the wrong track. The correct way of doing the question is to make the substitution $\displaystyle u = \frac{x \tan x}{1 + x}$. You therefore get $\displaystyle \int_a^b du = b - a$ where the integral terminals are left for you to find.

Originally Posted by Em Yeu Anh
Thanks for the help,
I was ill during the lectures that introduced integration, I'm having a lot of difficulties catching up.
Just to clarify the theory behind this, for a continuous function:
$\displaystyle \frac{d}{dx}\int_a^bf(t)dt = f(b) - f(a)$ is that what is happening in that question?

[snip]
This is wrong since the derivative of a definite integral with respect to the variable of integration is zero (because you're differentiating a constant). I apologise for putting you on the wrong track.

6. It's quite alright, I think I'm beginning to see it now:
If $\displaystyle u = \frac{xtanx}{1+x}$ then that upper limit will be $\displaystyle \frac{\frac{\pi}{4}}{\frac{5{\pi}}{4}} = \frac{1}{5}$ and the lower limit will be 0, leaving the final answer to be (1/5) - 0.