1
$\begingroup$

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.

  • 0
    I've added LaTeX formatting to your question; apologies if I changed your intended meaning in any way.2011-11-05
  • 3
    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.2011-11-05
  • 0
    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!2011-11-05
  • 2
    @JavaMan: I've removed the differentials (that weren't there before); see The Chaz's comment [here](http://meta.math.stackexchange.com/questions/3163/what-is-to-be-edited/3164#3164), the lack of differentials is part of what Jordan should be told about.2011-11-05
  • 0
    That's a great point. The differentials should in fact be removed. Thanks.2011-11-06
  • 0
    As an undergraduate who had to take all levels of Calculus - may I recommend the Calculus Lifesaver for your future studies, it gives a nice clear explanation I found for studying this material.2011-11-06
  • 0
    Someone here needs to learn to write things like \int f(x)\;dx\tag{6}.2011-11-06
  • 0
    I went to tutors, talking to the instructor doesn't really seem to help as much as going to tutors so I just do that instead.2011-11-06
  • 0
    @MichaelHardy: I assume your comment was directed at me. You are free to change my TeX code anytime you want. Just know that I am always willing to improve my TeX abilities, and there is no reason to be rude.2011-11-06
  • 0
    @JavaMan He was probably talking to me since I didn't know how to do that and he decided to be a child about it instead of acting like an adult.2011-11-06

3 Answers 3

2

$$ \frac{4+6u}{\sqrt{u}} = \frac{4}{\sqrt{u}} + \frac{6u}{\sqrt{u}} = 4u^{-1/2} + \frac{6\sqrt{u}\sqrt{u}}{\sqrt{u}} = 4u^{-1/2} + 6\sqrt{u} = 4u^{-1/2} + 6u^{1/2}. $$ 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.
  • 0
    $x\root3\of3\ne x^{1+1/3}$2011-11-06
  • 0
    @Gerry: The original problem was with $\sqrt[3]{x}$; seems like just a typo.2011-11-06
1

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?

  • 0
    I have no idea.2011-11-07
  • 1
    For the first, you'd get $F(b) - F(a)$ and the second you'd get ...? And how are they related?2011-11-07
  • 0
    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.2011-11-07
1

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.

  • 0
    I don't understand, it is a definite itegral so it should be a number?2011-11-07
  • 0
    @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.2011-11-07