How do you use the second fundamental theorem of Calculus to find the derivative of given #int cos(t) / t dt # from #[3, sqrtx]#?

1 Answer
Jul 22, 2017

Answer:

# d/dx \ int_3^(sqrt(x)) \ cost/t \ dt = cossqrt(x)/(2x) #

Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

# d/dx \ int_a^x \ f(t) \ dt = f(x) # for any constant #a#

(ie the derivative of an integral gives us the original function back).

We are asked to find:

# d/dx \ = int_3^(sqrt(x)) \ cost/t \ dt# ..... [A}

(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using a substitution and the chain rule. Let:

# u=sqrt(x) => (du)/dx = 1/2x^(-1/2) = 1/(2sqrt(x)) #

The substituting into the integral [A], and applying the chain rule, we get:

# d/dx \ int_3^(sqrt(x)) \ cost/t \ dt = d/dx \ int_3^u \ cost/t \ dt #

# " " = (du)/dx*d/(du) \ int_3^u \ cost/t \ dt #

# " " = 1/(2sqrt(x)) \ d/(du) \ int_3^u \ cost/t \ dt #

And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:

# d/dx \ int_3^(sqrt(x)) \ cost/t \ dt = 1/(2sqrt(x)) \ cosu/u #

And restoring the initial substitution we get:

# d/dx \ int_3^(sqrt(x)) \ cost/t \ dt = 1/(2sqrt(x)) \ cossqrt(x)/sqrt(x) #
# " " = cossqrt(x)/(2x) #