# Thread: Approximating definite integral using series

1. ## Approximating definite integral using series

Use series to approximate the definite integral I. (correct to 3 decimal places)

$I = \int^{1}_{0} x cos(x^5) dx$

Did I do my series right?

$\sum^{\infty}_{n=0} x (-1)^n \frac{(x)^{10n}}{(2n)!}$

2. Originally Posted by mollymcf2009
Use series to approximate the definite integral I. (correct to 3 decimal places)

$I = \int^{1}_{0} x cos(x^5) dx$

Did I do my series right?

$\sum^{\infty}_{n=0} x (-1)^n \frac{(x)^{10n}}{(2n)!}$
$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...$

$\cos(x^5) = 1 - \frac{x^{10}}{2!} + \frac{x^{20}}{4!} - \frac{x^{30}}{6!} + ...$

$x\cos(x^5) = x - \frac{x^{11}}{2!} + \frac{x^{21}}{4!} - \frac{x^{31}}{6!} + ...$

integrate ...

$\int_0^1 x\cos(x^5) \, dx = \left[\frac{x^2}{2} - \frac{x^{12}}{12 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!} - \frac{x^{32}}{32 \cdot 6!} + ... \right]_0^1
$

to be accurate to 3 decimal places ...

$\int_0^1 x\cos(x^5) \, dx \approx \left[\frac{x^2}{2} - \frac{x^{12}}{12 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!}\right]_0^1 = 0.460
$

3. Originally Posted by skeeter
$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...$

$\cos(x^5) = 1 - \frac{x^{10}}{2!} + \frac{x^{20}}{4!} - \frac{x^{30}}{6!} + ...$

$x\cos(x^5) = x - \frac{x^{11}}{2!} + \frac{x^{21}}{4!} - \frac{x^{31}}{6!} + ...$

integrate ...

$\int_0^1 x\cos(x^5) \, dx = \left[\frac{x^2}{2} - \frac{x^{12}}{12 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!} - \frac{x^{32}}{32 \cdot 6!} + ... \right]_0^1
$

to be accurate to 3 decimal places ...

$\int_0^1 x\cos(x^5) \, dx \approx \left[\frac{x^2}{2} - \frac{x^{12}}{12 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!}\right]_0^1 = 0.460
$
Awesome! I got exactly what you did, EXCEPT I forgot to add in the extra x before I integrated, so when I integrated, I was getting my exponents one less than I should have.

Thanks Skeeter!!