Maclaurin Series: use series for sin(5x^2) to evaluate int_0^0.77 sin(5x^2) dx

[Ezuras]

Assume that sin(x) equals its Maclaurin series for all x.

Use the Maclaurin series for sin(5x²) to evaluate the integral of sin(5x²) from 0 (bottom value on integral) to 0.77 (top).

Your answer will be an infinite series. Use the first two terms to estimate its value.

 

[Deleted member 4993]

Assume that sin(x) equals its Maclaurin series for all x.

Use the Maclaurin series for sin(5x²) to evaluate the integral of sin(5x²) from 0 (bottom value on integral) to 0.77 (top).

Your answer will be an infinite series. Use the first two terms to estimate its value.

What are your thoughts?

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[stapel]

Assume that sin(x) equals its Maclaurin series for all x.

So... what is the series for sin(x)?

Use the Maclaurin series for sin(5x²)....

So... what is the series for sin(5x2)?

...to evaluate the integral of sin(5x²) from 0 (bottom value on integral) to 0.77 (top).

Did they really give you an upper limit of "0.77"? Rather than, say, pi/4, or some other exact value?

Either way, what integral did you set up? How did you "use the Maclaurin series...to evaluate" that integral? Where are you getting stuck?

Please be complete. Thank you!