2 times the integral from 0 to 1 of (2t-3t^2) *sqrt(1-9t^2) dt.
Thank you very much.
First let's get that radicand simplified a bit:
Let
The first integral may be done using a simple substitution. I'll let you deal with that. For the second integral:
Take a moment and look at the graph I've posted below. You can easily see that this integrand does not exist beyond x = 1. So
<-- We need to do this, else the substitution below is illegal.
Now Let
I'll let you do this one also and then collect all the pieces.
-Dan
Quote:
Originally Posted by kittycat
what about for this integral,
the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much.
Instead of using the sine function, use the tangent function. (.)
-Dan
Hi Dan,
Could you please show me how to let the tangent function in this question? Sorry to ask you because I forgot the trig substitution. Thank you very much.
he same way that Soroban did, though in this case there are no restrictions due to the limits. So
Let
So
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Referring to : the integral from 0 to 1 of t^2 *sqrt(1+9t^2) dt.
I got to the same answer as you posted. But how should i evaluate it if it is =>the integral from 0 to 1 of t^2 *sqrt(1+9t^2) dt.
Could you please guide me to the final answer of this integration?
Thank you very much.