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Using the four non-zero terms of Maclaurin Series how can I find the value of $$\int_0^1 2x \cos ^2 x dx$$

When I solved I got Maclaurin Series as $2x-2x^3+\frac{2}{3}x^5$.

But I don't know how to apply the intervals.

Can anyone show how to do this.

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  • $\begingroup$ Please format your questions with MathJax so we can read them easily. $\endgroup$ – saulspatz Apr 7 '18 at 20:08
  • $\begingroup$ I don't think Maclaurin series are the right approach to compute that integral. $\endgroup$ – J.G. Apr 7 '18 at 20:08
  • $\begingroup$ But, I need to solve this using maclaurin series only. I have done this far and I don't know how to app those intervals. $\endgroup$ – sam Apr 7 '18 at 20:15
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First note that the first 4 (four) terms are $$2 x-2 x^3+\frac{2}{3} x^5 - \frac{4}{45} x^7$$ and the integral from this approximation is $\frac{3}{5}$. This is not too bad compared to the exact value $$\cos(1) \sin(1)+\frac{1}{2}\cos(1)^2 \approx 0.600612004276\approx 0.6 = \frac{3}{5}$$

Edit: An antiderivative of the four term expression is $$F(x) = x^2-\frac{1}{2}x^4+\frac{1}{9}x^6-\frac{1}{90}x^8$$ and therefore you get the integal as $F(1)-F(0) = \frac{3}{5} - 0$.

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  • $\begingroup$ Can you please explain it elaborately about how to use upper and lower limits. $\endgroup$ – sam Apr 7 '18 at 20:21
  • $\begingroup$ @sam: See the edit. $\endgroup$ – gammatester Apr 7 '18 at 20:27
  • $\begingroup$ So, if upper limit is 2, then I just need to substitute 2 in place of x. right $\endgroup$ – sam Apr 7 '18 at 20:34
  • $\begingroup$ Yes, that's right. But note the approximation is getting worse and worse if you increase the upper limit. $\endgroup$ – gammatester Apr 7 '18 at 20:38
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The right term should be approximation.

After you found the four non-zero terms,

$$2x\cos^2(x) \approx \sum_{i=1}^7 a_i x^i$$

Integrate it term by term.

$$\int_0^12x\cos^2(x)\, dx \approx \sum_{i=1}^7 a_i \int_0^1x^i\, dx$$

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$I =\int_0^12x\cos^2(x)\,dx$

The MacLaurin series of $f(x) = 2x\cos^2(x) $ is $2x - 2x^3 +\frac23x^5-\frac4{45}x^7 + \frac2{315}x^9........$

$I \approx \int_0^12x - 2x^3 +\frac23x^5-\frac4{45}x^7 \,dx$

$I\approx x^2-\frac12x^4 +\frac19x^6-\frac1{90}x^8\bigg]_0^1$

$I\approx 1-\frac12+\frac19-\frac1{90}$

$I \approx \frac35$

NOTE: im not sure if MacLaurin series are the best way to solve this integral ,I'd much rather use integration by parts

$I = (\frac{2x (\sin(2x)+x)+\cos(2x)}4)\bigg]_0^1$

$I = \frac{2.(\sin(2)+1)+\cos(2)}{4}-\frac{\cos(0)}4$

$I=0.600612$

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  • $\begingroup$ Double-check your integration by parts. $\endgroup$ – gammatester Apr 7 '18 at 20:30
  • $\begingroup$ @gammatester thank you ,i had taken 2 in degrees rather than radians. the rest was proper. $\endgroup$ – The Integrator Apr 8 '18 at 7:13

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