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Need help in finding the integral values:

  1. $$\int\limits (2x^4 + x^3/3 + \sqrt{x})\mathrm{d}x$$
  2. $$\int\limits_{\pi/2}^{3\pi/2} \cos(x)\mathrm{d}x$$
  3. $$\frac{\mathrm{d}}{\mathrm{d}x}\int\limits_{1}^{\tan(x)} e^t\mathrm{d}t$$ with the usage of the Fundamental Theorem of Calculus.
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up vote 2 down vote accepted
  1. $$\int(2x^4 + x^3/3 + \sqrt{x})\mathrm{d}x=\int2x^4 \mathrm{d}x+ \int \frac{x^3}{3}\mathrm{d}x +\int \sqrt{x}\mathrm{d}x=2\frac1{4+1}x^{4+1}+\frac{1}{3}\frac1{3+1}x^{3+1}+\frac1{\frac{1}{2}+1}x^{\frac{1}{2}+1}=\frac{2}{5}x^5+\frac{1}{12}x^4+\frac{2}{3}\sqrt[3]{x^2}+c $$
  2. $$\int\limits_{\pi/2}^{3\pi/2} \cos(x)\mathrm{d}x= \sin(x)|_{\pi/2}^{3\pi/2}=\sin\frac{3\pi}{2}-\sin \frac{\pi}{2}=-2$$
  3. By the 1st fundumental theorem of calculus and the chain rule if $\tan x=u$, $$\frac{\mathrm{d}}{\mathrm{d}x}\int\limits_{1}^{\tan(x)} e^t\mathrm{d}t= \frac{\mathrm{d}}{\mathrm{d}u}\int\limits_{1}^{u} e^t\mathrm{d}t\cdot \frac{\mathrm{d}u}{\mathrm{d}x}=e^u\frac{\mathrm{d}\tan x}{\mathrm{d}x}=e^{\tan x}\frac{1}{\cos^2 x} $$ I don't think one can be more thorough than this
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thank u sooo much, what theorems or applications did you use? – lias Dec 9 '12 at 11:25
For the first integral I used the linearity of the integral (first step) and then the fact that $\int x^n dx\frac{x^{n+1}}{n+1}$ for $n\neq -1$. For the second I used the fact that $\int \cos xdx=\sin x+c$ and for the 3rd I used the FTC and the Chain Rule – Nameless Dec 9 '12 at 11:27

Hint: $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}+C$ for any real number $n\neq -1$. Note also that integration is linear in the sense that $\displaystyle\int(f(x)+g(x))dx=\int f(x)dx+\int g(x)dx$. Therefore, we have $$\int\limits (2x^4 + x^3/3 + \sqrt{x})dx=2\int x^4dx+\frac{1}{3}\int x^3dx+\int x^{\frac{1}{2}}dx.$$ I think you can finish from here.

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i dont know how to apply this – lias Dec 9 '12 at 10:57
thank you, this is very helpful – lias Dec 9 '12 at 11:27

Fundamental theorem of Calculus (Question 3) $$ \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x) $$ Simply apply the above theorem with your $f(t)=e^t,h(x)=1,g(x)=\tan x$

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