Calculus question
Posted: Sun Sep 07, 2008 3:15 pm
If f(x)=\int_0^x{x^2sin(t^2)dt}, then what is f'(x)?
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Can we use the formulaIf f(x)=\int_0^x{x^2sin(t^2)dt}, then what is f'(x)?
the arc length l=int_0^a(sqrt[x'(t)^2+[y'(t)^2])dtI have not solved this. Here is a simpler one, but with a good idea.
Let a curve be defined parametrically by
x=\int_1^t{\frac{cos u}{u}du},
y=\int_1^t{\frac{sin u}{u}du}.
Find the arc length from the origin to the nearest point on the curve with a vertical tangent.
Not when there is a function of x in the integrand.Can we use the formula
f(x)=\int_a(x)^b(x){g(t)dt}------>f'(x)=g[b(x)]b'(x)-g[a(x)]a'(x)
Very close, but one detail needs to be changed.the arc length l=int_0^a(sqrt[x'(t)^2+[y'(t)^2])dt
if x=x(t)
y=y(t) then y'(x)=dy/dx=y'(t)/x'(t)----------> from this equation solve for a
and calculate the integral
Use the Leibnitz's rule :http://en.wikipedia.org/wiki/Leibniz_integral_ruleIf f(x)=\int_0^x{x^2sin(t^2)dt}, then what is f'(x)?