# Thread: I finally get how to find definite integrals but I keep getting this wrong!

1. ## I finally get how to find definite integrals but I keep getting this wrong!

I tried this twice and revised it as well but I don't get the correct answer. The correct answer is apparently 3.75 and I get 1.25. My work is attached as #23. I was hoping someone could point out my mistake. (I don't need full steps for what I did correctly)

Any help would be GREATLY appreciated!

2. Dear s3a,

$\displaystyle \int_{1}^{2}x^3dx$

$\displaystyle =\left[\frac{x^4}{4}\right]_{1}^{2}$

$\displaystyle =\frac{16}{4}-\frac{1}{4}$

$\displaystyle =\frac{15}{4}$

$\displaystyle =3.75$

I don't know the method of integration you had used. (I would much appreciate if you could briefly explain it to me.) But from the above method we see that the answer is clearly 3.75

Hope this helps.

3. Originally Posted by s3a
I tried this twice and revised it as well but I don't get the correct answer. The correct answer is apparently 3.75 and I get 1.25. My work is attached as #23. I was hoping someone could point out my mistake. (I don't need full steps for what I did correctly)

Any help would be GREATLY appreciated!
Hi s3a,

you have incorrectly expanded $\displaystyle (1+\frac{i}{n})^3$ on the 2nd line of no.23.

4. I have to go to bed now so I'll explain mine next time I visit this thread but I don't get your method either. Could you please elaborate it a bit for when I re-attempt this problem tomorrow? I am already lost at the first step that has the equal sign on the left of it (technically the second step).

5. $\displaystyle (1+\frac{i}{n})(1+\frac{i}{n})(1+\frac{i}{n})=(1+\ frac{2i}{n}+\frac{i^2}{n^2})(1+\frac{i}{n})$

$\displaystyle =1+\frac{i}{n}+\frac{2i}{n}+\frac{2i^2}{n^2}+\frac {i^2}{n^2}+\frac{i^3}{n^3}$

$\displaystyle =1+\frac{3i}{n}+\frac{3i^2}{n^2}+\frac{i^3}{n^3}$

6. Dear s3a,

Originally Posted by s3a
I have to go to bed now so I'll explain mine next time I visit this thread but I don't get your method either. Could you please elaborate it a bit for when I re-attempt this problem tomorrow? I am already lost at the first step that has the equal sign on the left of it (technically the second step).
Now you may know that, $\displaystyle \frac{d}{dx}x^{n+1}=(n+1)\frac{dx^{n}}{dx}$

Therefore, $\displaystyle \int{x^ndx}=\frac{x^{n+1}}{n+1}+C$ ; Where C is an arbitary constant.

Do you need any more clarifications? Please don't hesitate to reply me.

When you wish to evaluate an integral (or anti derivative) you take the following steps.

$\displaystyle \int x^3 dx$

first you start by using the power rule of integration which states simply that

$\displaystyle \int x^n = x^{n+1} / n + 1$

so in your case you have $\displaystyle \int x^3$

following the power rule gets you $\displaystyle x^4 /4$

now you need to evaluate it at your upper bound of 2

2^4 = 16 and 16/4 is equal to 4... thats your upper bown answer

now you need to find your lower bound solution... it's at 1

1^4 = 1 and 1/4 = .25 and thats your lower bound answer

so you take your upper bound solution of 4 and subtract from it your lower bound answer of .25 to get the correct answer of 3.75

the notation he's using in the "second" step of his post reads as follows

the function x^4 / 4 evaluated from 2 to 1

you evauate at top number (upper bound) and subtract from that the answer you get using the lower number (lower bound) as these match exactly from the intergrand symbol

8. Ok but with the way I was doing it before, I looked at my mistake with the cubic and I attempted soooooo many times to redo it and I guess I'm just keeping on making mistakes because of the repetion but can someone please spot my mistake this time without telling me to use another method. Before I consider anything else, I just want to get my method right.

9. Hi s3a,

you were doing very well, until the 4th line from the end

$\displaystyle [n+\frac{3(n+1)}{2}+\frac{3}{n^2}\frac{n(n+1)(2n+1) }{6}+\frac{1}{n^3}\frac{n^2(n+1)^2}{4}](\frac{1}{n})$

$\displaystyle =[n+\frac{3n+3}{2}+\frac{(3n+3)(2n+1)}{6n}+\frac{n^2 +2n+1}{4n}](\frac{1}{n})$

This is your only error now,