Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$\int_0^1(1 - x)^9\mathrm dx$$

$$\int_1^9\dfrac{\sqrt{u} - 2u^2}{u}\mathrm du$$

If someone could give me a clear step by step on these two I would appreciate it. I know they're simple but I can't find an example that explains the process clearly to me.


share|cite|improve this question
up vote 5 down vote accepted

These seem to be problems in finding the proper substitution. These problems fall under the category of "set the variable of substitution to be a complicated part".

For $\int_0^1(1 - x)^9 dx$, try $t=1-x$.

For $\int_1^9\dfrac{\sqrt{u} - 2u^2}{u} du$, try $t=\sqrt{u}$.

share|cite|improve this answer
the second one it might be easier to write $\frac{\sqrt{u}-2u^2}{u}=u^{-1/2}-2u $ – ShawnD Aug 23 '11 at 23:36
Thank you. This helped me get the first one, but I found the second one without substitution. Also thanks to @Shawn for that tip, it helped a lot. – ranonk Aug 24 '11 at 0:21
Definitely the method used for the second integral in this answer is needlessly complicated. – Michael Hardy Aug 24 '11 at 2:01
$\int_1^3\frac{t-2t^4}{t^2}\;2t\;\mathrm{d}t=\int_1^3(2-4t^3)\;\mathrm{d}t$ didnn't look that complicated; however, the problem can certainly be done without substitution. – robjohn Aug 24 '11 at 5:35
@robjohn see the deleted message – user 1618033 Dec 29 '14 at 21:18

When one first starts talking integration in (most standard) calculus class, there are only a small number of integrals which we evaluate. The functions with which we start discussing integration (are not limited to but) usually include:

  • $x^n$
  • $e^x$
  • $\sin x$
  • $\cos x$
  • $\tan x \sec x$
  • $\sec^2 x$
  • $-\csc^2 x$
  • $-\csc x \cot x$

(as time goes on, we of course start evaluating more complicated integrals). Remember also that we can easily integrate the sums of (two or more) well known functions, and we can easily integrate a constant times a well known function:

$$ \int(f(x) + g(x))dx = \int f(x) dx + \int g(x) dx $$ and $$ \int cf(x) dx = c \int f(x) dx. $$

Now, substitution is all about recognizing a more complicated integral as a composition of well known functions. When looking at the first integral, the integrad (that is, the function which is being integrated) is $(1-x)^9$. Of all the functions on the list, which function does it most resemble? It certainly most resembles $x^n$. So how do we make it of the form $u^n$ for some $u$? Simple: We let $u = 1-x$. This forces $du = - dx$ or $ - du = dx$. Plug everything back into the integral (remembering to change the bounds of integration), and we have the proper substitution.

The second integral looks more complicated. It's not, however. Remembering that we can add fractions if they have a common denominator:

$$ \frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}, $$

we can also split fractions in the same way:

$$ \frac{a+b}{c} = \frac{a}{c} + \frac{b}{c} $$

(it's the same process after all!). Recall also that the square root is really a power ($\sqrt x = x^{\frac{1}{2}}$). Now, try writing your second integral as a sum of two fractions, and manipulate the expression (using algebra) to reduce it to something you know.

share|cite|improve this answer
Thank you for your response. I feel like I understand integration better now. – ranonk Aug 24 '11 at 0:22

$$ \int_1^9 \frac{\sqrt{u} - 2u^2}{u}\;du = \int_1^9 \frac{\sqrt{u}}{u}\;du - \int_1^9 \frac{2u^2}{u}\;du = \int_1^9 u^{-1/2}\;du - \int_1^9 2u\;du. $$ Do the two separately.

share|cite|improve this answer

$$\begin{align*} \int^1_0 {(1-x)^9} dx & = G(1) - G(0) \\ G'(x) & = \int {(1-x)^9} dx \\ G(x) & = \frac{(1-x)^{10}}{10} \\ G(1) & = \frac{(1-1)^{10}}{10} = 0 \\ G(0) & = \frac{(1-0)^{10}}{10} = \frac{1}{10} \\ G(1) - G(0) & = -\frac{1}{10}\end{align*}$$

share|cite|improve this answer
I fixed your LaTeX for the sake of readability and some typos. If you click on edited xx time ago above my name you can see what I did. – t.b. Aug 25 '11 at 13:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.