Let F(x)= x^2
∫ sqrt(1+t^3)dt
x
Find F'(4).
I applied the FTC and thought that F'(4) = f(4) which in this case would be sqrt(1+4^3), giving me an answer of sqrt(65), however this is incorrect. I'm not sure how to proceed with the question.
2 answers
That x^2 should be the upper limit on the integral and x is the lower limit on the integral
You forgot the chain rule.
F'(x) = √(1+(x^2)^3)*(2x) - √(1+x^3)*(1)
F'(4) = √(1+4096)(8)-√(1+64)
visit this url and scroll down about halfway -- there is a good example with variable limits of integration.
https://en.wikipedia.org/wiki/Leibniz_integral_rule
F'(x) = √(1+(x^2)^3)*(2x) - √(1+x^3)*(1)
F'(4) = √(1+4096)(8)-√(1+64)
visit this url and scroll down about halfway -- there is a good example with variable limits of integration.
https://en.wikipedia.org/wiki/Leibniz_integral_rule
7 answers