How do you evaluate #d /dx \int_[x]^{x^4} sqrt{t^2+t} dt#?
2 Answers
To solve
**don't forget to chain!!
# d/dx \ int_x^(x^4) \ sqrt(t^2+t) \ dt = 4x^3sqrt(x^8+x^4) - sqrt(x^2+x) #
Explanation:
If asked to find the derivative of an integral using the fundamental theorem of Calculus, you should not evaluate the integral.
The Fundamental Theorem of Calculus tells us that:
# d/dx \ int_a^x \ f(t) \ dt = f(x) #
(ie the derivative of an integral gives us the original function back).
We are asked to find:
# d/dx \ int_x^(x^4) \ sqrt(t^2+t) \ dt #
(notice the upper and lower bounds are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral as follows:
# int_x^(x^4) \ sqrt(t^2+t) \ dt = int_0^(x^4) \ sqrt(t^2+t) - int_0^(x) \ sqrt(t^2+t) \ dt#
We have arbitrary chosen the lower limit as
# d/dx \ int_0^(x) \ sqrt(t^2+t) \ dt = sqrt(x^2+x) #
And using the chain rule we can write:
# d/dx int_0^(x^4) \ sqrt(t^2+t) = (d(x^4))/dx d/(d(x^4)) int_0^(x^4) \ sqrt(t^2+t) #
Now,
# d/(d(x^4)) int_0^(x^4) \ sqrt(t^2+t) = sqrt((x^4)^2+(x^4))#
Hence combining these trivial results we get:
# d/dx \ int_x^(x^4) \ sqrt(t^2+t) \ dt = 4x^3sqrt(x^8+x^4) - sqrt(x^2+x) #