Estimating v power series

We have the right to use power collection to calculation definite integrals in the same way we provided them to calculation indefinite integrals. The only distinction is the we’ll evaluate end the given interval as soon as we uncover a power series that to represent the initial integral.

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To evaluate end the interval, we’ll broaden the power collection through that is first couple of terms, and also then evaluate each term independently over the interval.


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Oftentimes fine be request to use a power series to almost right the identify integral come a certain variety of decimal places. If this is the case, we should make certain we keep an ext decimals than we’re request for once we evaluate over the interval. That way, we’ll have the ability to give an accurate answer come the requested number of decimal places when we sum every one of our decimal values together.



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Estimating the identify integral to 5 decimal places using a usual power series

Example

Use power collection to calculation the identify integral to five decimal places.

???\int^0.2_04x\arctan(2x)\ dx???

Since this integral contains an ???\arctan??? function, we’ll use the typical power series

???\arctanx=\sum^\infty_n=0\frac(-1)^nx^2n+12n+1???

and manipulate the to obtain the given definite integral.

We can start by replacing ???x??? with ???2x??? inside the ???\arctan??? duty in stimulate to match the offered function.

???\arctan(2x)=\sum^\infty_n=0\frac(-1)^n(2x)^2n+12n+1???

Next, we’ll multiply both political parties by ???4x??? in bespeak to make the power series match the function.

???4x\arctan(2x)=4x\sum^\infty_n=0\frac(-1)^n(2x)^2n+12n+1???

???4x\arctan(2x)=4x^1\sum^\infty_n=0\frac(-1)^n2^2n+1x^2n+12n+1???

???4x\arctan(2x)=\sum^\infty_n=0\frac4(-1)^n2^2n+1x^2n+22n+1???

Since the problem asks us to find the integral that the left-hand next of this equation, we’ll integrate both sides.

???\int4x\arctan(2x)\ dx=\int\sum^\infty_n=0\frac4(-1)^n2^2n+1x^2n+22n+1\ dx???


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Because we want to settle for what we have actually on the left-hand side, we only require to incorporate the appropriate side. We’re integrating v respect to ???x???, for this reason we can treat the ???n???’s together constants.

???\int4x\arctan(2x)\ dx=\sum^\infty_n=0\frac4(-1)^n2^2n+1x^2n+3(2n+1)(2n+3)???

Adding in the interval native the original question, we get

???\int^0.2_04x\arctan(2x)\ dx=\left<\sum^\infty_n=0\frac4(-1)^n2^2n+1x^2n+3(2n+1)(2n+3)\right>\Bigg|^0.2_0???

In order to evaluate over the interval, we’ll broaden the power collection through the first few terms. Remember, we have to approximate the last answer to five decimal places, which method we’ll need to calculate results past five decimals until we obtain to a suggest where the an initial five decimal places aren’t changing.

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???\int^0.2_04x\arctan(2x)\ dx=\frac8x^33-\frac32x^515+\frac128x^735-\frac512x^963+\frac2,048x^1199+...\bigg|^0.2_0???

Evaluating each term individually over the interval, us get

???\int^0.2_04x\arctan(2x)\ dx=\left(\frac8(0.2)^33-\frac8(0)^33\right)-\left(\frac32(0.2)^515-\frac32(0)^515\right)???

???+\left(\frac128(0.2)^735-\frac128(0)^735\right)-\left(\frac512(0.2)^963-\frac512(0)^963\right)???

???+\left(\frac2,048(0.2)^1199-\frac2,048(0)^1199\right)+...???

???\int^0.2_04x\arctan(2x)\ dx\approx0.0213333-0.0006827+0.0000468-0.0000041+0.0000004+...???

Let’s start including the terms together.

???n_0+n_1=0.0206506???

???n_0+n_1+n_2=0.0206974???

???n_0+n_1+n_2+n_3=0.0206933???

???n_0+n_1+n_2+n_3+n_4=0.0206937???

Remember, us only require the an initial five decimal places. When we analysis our results, we can see that ???n_0+n_1+n_2+n_3=0.0206933??? is as far as we need to add in order to get 5 stable decimal points. Therefore, round off the price to five decimal places gives