How do you calculate the derivative of #intcos(t)/t dt# from #[x,3]#?

Question

Scarlett Allison · Accepted Answer

Use the Fundamental Theorem of Calculus Part 1.

Fundamental Theorem of Calculus Part 1 says (in part) that if #f# is continuous on #[a,b]# and function #g# is defined on #[a,b]# by #g(x) = int_a^x f(t) dt#, then #g'(x) = f(x)# (for all #x in (a,b)#) This question asks about the function: #g(x) = int_x^3 cos(t)/t dt# Clearly, in this question we have #f(t) = cot(t)/t#. Notice that FTC 1 requires the constant to be the lower limit of integration, so we use the properties of definite integral to write: #g(x) = - int_3^x cos(t)/t dt# Now, we can see that #g'(x) = - cos(x)/x#. That's it. Stop! We are finished. Move on to the next question. (On exams, the trickiest thing about this kind of question is often understanding how easy it is. We expect more work to be needed.) example Find the derivative of: #g(x) = int_1^x sin((2pit)/3)/sqrt(t^3-2t) dt# It is: #g'(x) = sin((2pix)/3)/sqrt(x^3-2x)#. (It is so easy, that graders may penalize heavily for forgetting to change the variable #t# to #x#.)

Charles Arocha · Answer

undefined To calculate the derivative of the integral of cos(t)/t from [x,3] with respect to x, you can use the Leibniz Rule, which states:

d/dx ∫[a(x), b(x)] f(t,x) dt = ∫[a(x), b(x)] ∂f/∂x dt + f(b(x), x) * db(x)/dx - f(a(x), x) * da(x)/dx

In this case, a(x) = x and b(x) = 3. So, applying the Leibniz Rule:

d/dx ∫[x,3] cos(t)/t dt = ∫[x,3] (∂/∂x (cos(t)/t)) dt + (cos(3)/3) * d(3)/dx - (cos(x)/x) * d(x)/dx

Now, compute the partial derivative of cos(t)/t with respect to x:

∂/∂x (cos(t)/t) = -sin(t)/t

Substitute back into the expression:

d/dx ∫[x,3] cos(t)/t dt = ∫[x,3] (-sin(t)/t) dt + (cos(3)/3) * 0 - (cos(x)/x) * 1

Integrate -sin(t)/t from x to 3:

= ∫[x,3] (-sin(t)/t) dt
= -Ci(3) + Ci(x) - π/2

Where Ci is the cosine integral function.

Therefore, the derivative of the integral of cos(t)/t from [x,3] with respect to x is:

= -Ci(3) + Ci(x) - π/2 - (cos(x)/x)