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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.

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    @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

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$ \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.
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    @Gerry: The original problem was with $\sqrt[3]{x}$; seems like just a typo.2011-11-06
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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?

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    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
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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.

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    @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