Estimate the area under the curve cos sqrt x for 0=
Definite Integrals? Any help..im stuck
Use trapezoidal rule. If you look at a plot of your function, you'll see that it falls roughly linearly from 0 to 2.
The answer with trap. rule would be about 1.156
the problem is we havnt got to trapeziod rule
Not sure why it's asking you to estimate when you can solve it exactly with a u sub then by parts.
\[\int\limits_{a}^{b} f(x)dx \approx (b-a)\frac{f(a)+f(b)}{2}\]
Well then you have to estimate with rectangles.
http://www.wolframalpha.com/input/?i=integral+cos%28sqrt%28x%29%29+from+x%3D0..2
You should know that the area of a trapezoid is \[\frac{1}{2}(a+b)h\]where a and b are the 'top' and 'bottom' and h is the altitude (or height). Here, a=f(0), b=f(2) and the height is (b-a)=2
It's okay to apply this formula.
thanks guys
np
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