Please help me integrate this
- Thread starter greese
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BigGlenntheHeavy
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\(\displaystyle \int_{0}^{1}x^{2}cos(x)dx\)
Used integration by parts twice (one way).
Used integration by parts twice (one way).
It appears they want you to integrate and show that the result is less than or equal to 1/3.
\(\displaystyle \int_{0}^{1}x^{2}cos(x)dx\leq \frac{1}{3}\)
You can use parts. Let \(\displaystyle u=x^{2}, \;\ dv=cos(x)dx, \;\ du=2xdx, \;\ v=sin(x)\)
\(\displaystyle x^{2}sin(x)-2\int xsin(x)dx\)
Do it again:
\(\displaystyle u=x, \;\ dv=sin(x)dx, \;\ du=dx, \;\ v=-cos(x)\)
Now, can you finish putting it all together?.
You could also do tabular integration.
It is related to parts.
Let \(\displaystyle u=x^{2}, \;\ dv=v'dx=cos(x)dx\)
Create a table consisting of three columns:
\(\displaystyle \begin{tabular}{ccccc}Alternate signs&{}&u and its derivatives&{}&v' and its derivatives\\+&\rightarrow&x^{2}&\searrow&cos(x)\\-&\rightarrow&2x&\searrow&-sin(x)\\+&\rightarrow&2&\searrow&-cos(x)\\-&\rightarrow&0&{}&sin(x)\\ {}&{}&\uparrow&{}&{}\\{}&{}&Differentiate until you get 0&{}&{}\end{tabular}\)
Now, multiply the second column by the term below the se arrow and then add them up. And don't forget the alternating signs.
\(\displaystyle \boxed{x^{2}sin(x)+2xcos(x)-2sin(x)}\)
I wanted to show you that method so you will know it for other problems similar to this. It works well if you have a trig function multiplied by an x term.
Just playing around with some LaTex graphics:
\(\displaystyle \setlength{\unitlength}{1mm}
\begin{picture}(60, 40)
\put(20, 30){\circle{1}}
\put(20, 30){\circle{2}}
\put(20, 30){\circle{4}}
\put(20, 30){\circle{8}}
\put(20, 30){\circle{16}}
\put(20, 30){\circle{32}}
\end{picture}\)
\(\displaystyle \int_{0}^{1}x^{2}cos(x)dx\leq \frac{1}{3}\)
You can use parts. Let \(\displaystyle u=x^{2}, \;\ dv=cos(x)dx, \;\ du=2xdx, \;\ v=sin(x)\)
\(\displaystyle x^{2}sin(x)-2\int xsin(x)dx\)
Do it again:
\(\displaystyle u=x, \;\ dv=sin(x)dx, \;\ du=dx, \;\ v=-cos(x)\)
Now, can you finish putting it all together?.
You could also do tabular integration.
It is related to parts.
Let \(\displaystyle u=x^{2}, \;\ dv=v'dx=cos(x)dx\)
Create a table consisting of three columns:
\(\displaystyle \begin{tabular}{ccccc}Alternate signs&{}&u and its derivatives&{}&v' and its derivatives\\+&\rightarrow&x^{2}&\searrow&cos(x)\\-&\rightarrow&2x&\searrow&-sin(x)\\+&\rightarrow&2&\searrow&-cos(x)\\-&\rightarrow&0&{}&sin(x)\\ {}&{}&\uparrow&{}&{}\\{}&{}&Differentiate until you get 0&{}&{}\end{tabular}\)
Now, multiply the second column by the term below the se arrow and then add them up. And don't forget the alternating signs.
\(\displaystyle \boxed{x^{2}sin(x)+2xcos(x)-2sin(x)}\)
I wanted to show you that method so you will know it for other problems similar to this. It works well if you have a trig function multiplied by an x term.
Just playing around with some LaTex graphics:
\(\displaystyle \setlength{\unitlength}{1mm}
\begin{picture}(60, 40)
\put(20, 30){\circle{1}}
\put(20, 30){\circle{2}}
\put(20, 30){\circle{4}}
\put(20, 30){\circle{8}}
\put(20, 30){\circle{16}}
\put(20, 30){\circle{32}}
\end{picture}\)
D
Deleted member 4993
Guest
greese said:
I think the real problem is to prove the inequality without actally doing the integration and finding the values (of sin(1) etc.)
What topic are you discussing in our class now?