How do you evaluate #d /dx \int_[x]^{x^4} sqrt{t^2+t} dt#?

Question

Paisley Johnson · Accepted Answer

# d/dx \ int_x^(x^4) \ sqrt(t^2+t) \ dt = 4x^3sqrt(x^8+x^4) - sqrt(x^2+x) #

You shouldn't evaluate an integral if you are asked to find its derivative using the calculus fundamental theorem. According to the Calculus Fundamental Theorem: # d/dx \ int_a^x \ f(t) \ dt = f(x) # (i.e., we can return to the original function using the derivative of an integral). It is requested of us to locate: # d/dx \ int_x^(x^4) \ sqrt(t^2+t) \ dt # We can manipulate the definite integral as follows (notice the upper and lower bounds are not in the correct format for the FTOC to be applied, directly): # 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 #0# wlog (any number will do!). The second integral is is now in the correct form, and we can directly apply the FTOC and write the derivative as: # d/dx \ int_0^(x) \ sqrt(t^2+t) \ dt = sqrt(x^2+x) # The chain rule also allows us to 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(x^4))/dx=4x^3#, And, using the FTOC, we have: # d/(d(x^4)) int_0^(x^4) \ sqrt(t^2+t) = sqrt((x^4)^2+(x^4))# When we combine these unimportant findings, we obtain: # d/dx \ int_x^(x^4) \ sqrt(t^2+t) \ dt = 4x^3sqrt(x^8+x^4) - sqrt(x^2+x) #

Scarlett Allison · Answer

To solve #d /dx \int_x^{x^4} sqrt{t^2+t} dt#, we will use the Fundamental Theorem of Calculus. #d /dx \int_x^{x^4} sqrt{t^2+t} dt#
#= [\sqrt{(x^4)^2+x^4}] * (4x^3) - [\sqrt{(x)^2+x}] * (1)# 
**don't forget to chain!!

Ezra Adams · Answer

undefined To evaluate $ \frac{d}{dx} \int_{x}^{x^4} \sqrt{t^2 + t} \, dt $, we apply the Fundamental Theorem of Calculus and the Chain Rule.

First, we find the antiderivative of $ \sqrt{t^2 + t} $ with respect to $ t $. Let $ F(t) $ be the antiderivative. Then,

$$ F(t) = \int \sqrt{t^2 + t} \, dt $$

Next, we evaluate $ F(x^4) - F(x) $.

$$ \frac{d}{dx} \int_{x}^{x^4} \sqrt{t^2 + t} \, dt = F'(x^4) \cdot (4x^3) - F'(x) \cdot 1 $$

This simplifies to:

$$ 4x^3 \cdot \sqrt{(x^4)^2 + x^4} - \sqrt{x^2 + x} $$

$$ = 4x^3 \cdot \sqrt{x^8 + x^4} - \sqrt{x^2 + x} $$