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

I have been trying to study for a test on monday but I can't do any of the basic problems. I know what to do but I am just not good enough at math to get the proper answer.

I am supposed to use part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. I am given integral $$\hskip1in \int_x^\pi\cos(\sqrt{t}) \; dt\tag{1}$$ I know that if it is from $\pi$ to $x$ all I do is replace $t$ with $x$ and I get the answer. There are no examples in this book of how to do this so I do not know how to do it.

Next I am supposed to evaluate the integrals and I got to a point in the problems where I can't do any of them after number 27 out of 45.

$$\int_0^\pi (5e^x + 3\sin(x))\tag{2}$$ I just can't think of how to get the integral for this.

$$\int_1^4\frac{4+6u}{\sqrt{u}}\tag{3}$$ I don't know substitution yet so I don't know how to do this one either. I know I could try and subtract exponent u's but the 4 has no u so it is not possible I think.

$$\int_0^1 x(\sqrt[3]{x} + \sqrt[4]{x}) \; dx\tag{4}$$ Again no idea what to do here, I made it into $x^{3/2} + x^{5/4}$ but that is still wrong after I integrate that.

$$\int_1^2\left(\frac{x}{2} - \frac{2}{x}\right)\;dx\tag{5}$$ I do not know what to do with this one either I tried to make it $\frac{1}{2x}$ and something else I think it might integrate to $\frac{1}{3}x^{3/2} - 2\ln(x)$ but I can't get a proper answer out of that.

$$\int_0^1 (x^{10} + 10^x)$$ I tried many things on this but was never right. $\frac{1}{11} x^{11} + 10x^{x+1}\tag{6}$ is not right but should be to me.

$$\int_0^{\pi/4} \frac{1+\cos^2(x)}{\cos^2(x)}\tag{7}$$ I tried $\frac{1}{\cos^2(x)}$ but to me that means nothing and I am left with maybe $\ln(\cos^2(x)) + 1$ but that is wrong.

share|cite|improve this question
I've added LaTeX formatting to your question; apologies if I changed your intended meaning in any way. – Zev Chonoles Nov 5 '11 at 23:42
There are a LOT of problems in here you seem to be stuck on, perhaps you could return to the previous several sections of your textbook/class notes and review the material? Most of them look pretty simple if you realize you can write the integrals in more simple terms using algebra. – Tyler Nov 5 '11 at 23:46
Seems like you really could use some help. Have you gone to your professor or whoever is teaching you and asked for advice. It looks like you have problems in basic algebra ($x/2$ is not the same as $x^{1/2}$) and knowledge of what an exponential is (if you write $10^x$ as $e^{x \ln(10)}$ can you integrate that?). In general, get help! – marty cohen Nov 5 '11 at 23:55
@JavaMan: I've removed the differentials (that weren't there before); see The Chaz's comment here, the lack of differentials is part of what Jordan should be told about. – Zev Chonoles Nov 5 '11 at 23:58
That's a great point. The differentials should in fact be removed. Thanks. – JavaMan Nov 6 '11 at 0:00
up vote 2 down vote accepted

$$ \frac{4+6u}{\sqrt{u}} = \frac{4}{\sqrt{u}} + \frac{6u}{\sqrt{u}} = 4u^{-1/2} + \frac{6\sqrt{u}\sqrt{u}}{\sqrt{u}} = 4u^{-1/2} + 6\sqrt{u} = 4u^{-1/2} + 6u^{1/2}. $$ Now deal with the two terms separately.

In (4), multiply first: $$ x\left(\sqrt[3]{x} + \sqrt[4]{x}\right) = x\sqrt[3]{x} + x\sqrt[4]{x} = x^{1+1/3} + x^{1+1/4} = x^{4/3} + x^{5/4}. $$ In (6): the power rule applies only when the exponent is constant.

In (7), you have two problems:

  • you should recognize that $1/\cos^2 x= \sec^2 x$;
  • you should not assume that $\displaystyle\int\dfrac{1}{\text{anything}}\;dx = \ln(\text{anything})+C$. It doesn't work that way.
share|cite|improve this answer
$x\root3\of3\ne x^{1+1/3}$ – Gerry Myerson Nov 6 '11 at 4:02
@Gerry: The original problem was with $\sqrt[3]{x}$; seems like just a typo. – Arturo Magidin Nov 6 '11 at 4:35

In (1), are you saying you don't know any relation between $$\int_a^bf(x)\,dx{\rm\ and\ }\int_b^af(x)\,dx$$ If so, suppose you knew $\int f(x)\,dx=F(x)+C$. Then what would you get for the two integrals in the display, and how would they be related?

share|cite|improve this answer
I have no idea. – user138246 Nov 7 '11 at 14:25
For the first, you'd get $F(b) - F(a)$ and the second you'd get ...? And how are they related? – Tyler Nov 7 '11 at 17:03
It sounds like you are saying that you don't know what the notation $\int_a^bf(x)\,dx$ means, and in particular you don't know how to evaluate it, ever. In that case, the first thing to do is to learn what the notation means. You don't have a snowball's chance in Hell of answering your other questions, if you don't know what a definite integral is. – Gerry Myerson Nov 7 '11 at 21:56

In (5), you have $$\int_1^2\left(\frac{x}{2} - \frac{2}{x}\right)\,dx.$$ This is: $$\begin{align*} \int_1^2\left(\frac{x}{2} - \frac{2}{x}\right)\,dx &= \int_1^2\left(\frac{1}{2}\;x - 2\;\frac{1}{x}\right)\,dx \\ &= \int_1^2\frac{1}{2}\;x\,dx - \int_1^22\;\frac{1}{x}\,dx\\ &= \frac{1}{2}\int_1^2x\,dx - 2\int_1^2\frac{1}{x}\,dx. \end{align*}$$ Now do each integral.

The integral cannot "integrate to $\frac{1}{3}x^{3/2}-2\ln x$" because this is a definite integral, so it should integrate to a number.

share|cite|improve this answer
I don't understand, it is a definite itegral so it should be a number? – user138246 Nov 7 '11 at 14:24
@Jordan: A definite integral, $\int_a^b f(x)\,dx$, is the net signed area between the $x$-axis, the graph of $y=(x)$, and the lines $x=a$ and $x=b$. Being a net signed area, it equals a number, not a function. – Arturo Magidin Nov 7 '11 at 14:48

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.