# Please check me on this integral! **URL:** \ **Category:** Factual Questions\ **Created:** [April 18, 2002, 5:22am UTC](https://boards.straightdope.com/t/please-check-me-on-this-integral/104271 "2002-04-18T05:22:54Z")\ **Posts on this page:** 6\ **Page:** 1
**Author:** ![moe.ron](https://avatars.discourse-cdn.com/v4/letter/m/ed8c4c/32.png) [@moe.ron](https://boards.straightdope.com/u/moe.ron)\ **Post date:** [April 18, 2002, 5:22am UTC](https://boards.straightdope.com/t/please-check-me-on-this-integral/104271/1 "2002-04-18T05:22:54Z")
So I have homework due for calculus. I acknowledge that most of you are smarter than I (for now…), so I humbly come to you seeking answers. I looked through a book of integrals to find this, work through it, and I really don’t think it’s right. Here’the OG problem: (integral from (-1) to (4)) of (2x)/ ((3x+4)^1/4)) is 8 \* ((x/3) - ((4/9)\* log (4+3x))) I get (9.0517). Would anyone care to check me? I’ve done it a few times and come up with the same answer, but I don’t feel very confident about it. Any positive renforcement would be greatly appreciated. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [April 18, 2002, 6:16am UTC](https://boards.straightdope.com/t/please-check-me-on-this-integral/104271/2 "2002-04-18T06:16:17Z")
> [@](#): > > \*Originally posted by moe.ron \* > \*\*So I have homework due for calculus. I acknowledge that most of you are smarter than I (for now…), so I humbly come to you seeking answers. I looked through a book of integrals to find this, work through it, and I really don’t think it’s right. Here’the OG problem: > > (integral from (-1) to (4)) of (2x)/ ((3x+4)^1/4)) is > > 8 \* ((x/3) - ((4/9)\* log (4+3x))) > > I get (9.0517). Would anyone care to check me? I’ve done it a few times and come up with the same answer, but I don’t feel very confident about it. Any positive renforcement would be greatly appreciated. \*\* My Mathcad program (Mathsoft Inc.) gives 7.831. The integral equals 2/15\*(3x+4)[sup]5/3[/sup]-4/3\*(3x+4)[sup]2/3[/sup] ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [April 18, 2002, 6:58am UTC](https://boards.straightdope.com/t/please-check-me-on-this-integral/104271/3 "2002-04-18T06:58:35Z")
My hand-held calculator (Texas Instruments) concurs with **David Simmons** on the value (1480/189 = 7.83069) but not the integrated fuction. It gets: (8/189)(3x+4)[sup]3/4/sup Odd. When I plug in the limits on your function, **David** , I get something completely different—6.27968. ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [April 18, 2002, 7:35am UTC](https://boards.straightdope.com/t/please-check-me-on-this-integral/104271/4 "2002-04-18T07:35:28Z")
To do it by hand, try integrating by parts with u = 2x and dv = dx / [3x+4][sup]1/4[/sup] I think **David** had a typo when he was finding the antiderivative (maybe ^(1/3) instead of ^(1/4)?), but both he and **Achernar** have the definite integral correct. ( **Achernar** ’s antiderivative is correct, but that may not be immediately obvious from what you get when you do it by hand). ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [April 18, 2002, 3:16pm UTC](https://boards.straightdope.com/t/please-check-me-on-this-integral/104271/5 "2002-04-18T15:16:37Z")
Good call, **Cabbage**. When I do the integral by hand, I get (unsimplified): 8 / 63 × (3x + 4)[sup]7/4[/sup] - 32 / 27 × (3x + 4)[sup]3/4[/sup] However, I don’t think it’s necessary to solve the indefinite integral as stated. I think that the best way to do this problem is with u-substitution, with u = (3x+4)[sup]1/4[/sup]. It really de-uglifies the integrand. The fact that the limits on u work out so well if you do this suggests to me that this is a textbook integral which was intended to be solved with u-substitution. Integration by parts works fine, too, but I personally don’t like integrating fractional powers. 🙂 And if they’re still teaching u-substitution before integration by parts, **moe.ron** may not be able to take that route. As a general hint for solving integrals, one of your best bets is to differentiate your final result, and see if you can make it look like your original integrand. Also, a handy resource for indefinite integrals can be found at [integrals.com](http://integrals.com/). ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [April 18, 2002, 4:16pm UTC](https://boards.straightdope.com/t/please-check-me-on-this-integral/104271/6 "2002-04-18T16:16:23Z")
> [@](#): > > \*Originally posted by Achernar \* > \*\*Good call, **Cabbage**. When I do the integral by hand, I get (unsimplified): > > 8 / 63 × (3x + 4)[sup]7/4[/sup] - 32 / 27 × (3x + 4)[sup]3/4[/sup] > > \*\* If I put the correct function into Mathcad, I get the above answer so I guess we are all in agreement. I must have entered the wrong function under the integral sign.