# Estimating integral with Maclaurin series

## Homework Statement

Assume that sin(x) equals its Maclaurin series for all x. Use the first two terms of the Maclaurin series for sin(7x^2) to approximate the integral:

$$\int_{0}^{0.77}sin(7x^{2})\ dx$$

## The Attempt at a Solution

If I understand correctly, a Maclaurin series is just a Taylor series centered at 0. Therefore, my first two terms would be:

$$sin(7(0)^{2})$$

and

$$14(0)^{2}cos(7(0)^{2})$$

I feel like I'm getting off on the wrong foot right away.

Help anyone?

your first derivative is slightly off it should just 14xcos(7x^2), evaluated at zero. which just equals zero.

## Homework Statement

If I understand correctly, a Maclaurin series is just a Taylor series centered at 0. Therefore, my first two terms would be:

$$sin(7(0)^{2})$$

and

$$14(0)^{2}cos(7(0)^{2})$$

I feel like I'm getting off on the wrong foot right away.

Help anyone?

Yes you are doing this correctly so far except your second term should be

$$14(x) cos(7(x)^{2}) \right _{x=0}$$

which of course gives the same result. But this could have messed you up with your next derivatives.

Now try to take a few more derivatives until you get some non zero terms. Don't forget about the product rule

The original problem statement says to use the first two terms of the Maclaurin series. So when I am constructing the series, I just basically throw away any terms with zero values, and the non-zero terms then become my series.

Is this correct?

SammyS
Staff Emeritus
Homework Helper
Gold Member
One way to do this is to take the Maclaurin series for sin(x), then substitute 7x2 for x. Then integrate that tern by term.

The original problem statement says to use the first two terms of the Maclaurin series. So when I am constructing the series, I just basically throw away any terms with zero values, and the non-zero terms then become my series.

Is this correct?

Yes, just use the first two non zero terms.

SammyS, your idea sounds MUCH simpler, but you would have to explain the details of implementation to me.

So, doing it the hard way - and yes, it gets a bit messy, so I won't post all the details.

I end up with the first two non-zero terms being term 2, which evaluates to 14, and term 6, which evaluates to -41160.

Dividing these with their factorials, multiplying by the x-a terms (which is just x, as a is zero) and simplifying, I end up with:

$$7x^{2}-\frac{343\ x^{6}}{6}$$

Plugging in 0.77, I get a result of -7.764512728421

I know the actual value of the integral is about 0.3

I presume the error comes from the fact that I am only using two terms.

Am I correct?

Last edited:
Char. Limit
Gold Member
SammyS, your idea sounds MUCH simpler, but you would have to explain the details of implementation to me.

So, doing it the hard way - and yes, it gets a bit messy, so I won't post all the details.

I end up with the first two non-zero terms being term 2, which evaluates to 14, and term 5, which evaluates to -41160.

Dividing these with their factorials, multiplying by the x-a terms (which is just x, as a is zero) and simplifying, I end up with:

$$7x^{2}-\frac{343\ x^{6}}{6}$$

Plugging in 0.77, I get a result of -7.764512728421

I know the actual value of the integral is about 0.3

I presume the error comes from the fact that I am only using two terms.

Am I correct?

But you have to integrate it first. You can't just plug in .77 before you integrate.

Okay, yep, missed that one completely.

Integrating gives me:

$$\frac{7x^{3}}{3}-\frac{49x^{7}}{6}$$

Now, plugging in 0.77, I get -0.245385 which is much closer.

Again, I presume the error is from only using two terms.

Char. Limit
Gold Member
Yes it is. If you had tried three terms, you would get a number even close, but above the correct answer.

Thank you all for your help.

Char. Limit
Gold Member
Have a great day!