Find the derivative of #int_0^(x^3e^x) \(t^3+3)^17 \ dt#?

Question

Christopher Allers · Accepted Answer

Because we need the chain rule.

#y = g(u) = int_0^u (t^3+3)^17 dt# has #dy/(du) = (u^3+3)^17# and #dy/dx = dy/(du) * (du)/dx# so #g'(x) = dy/dx = (u^3+3)^17 * d/dx(x^3e^3)#

Samantha Adkison · Answer

# d/dx int_0^(x^3e^x) \ (t^3+3)^17 \ dt = (x^3e^x + 3x^2e^x)((x^3e^x)^3+3)^17#

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_0^(x^3e^x) \(t^3+3)^17 \ 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 the product rule, and the chain rule. Let: # u = x^3e^x => (du)/dx = (x^3)(d/de^x) + (d/dxx^3)(e^x) # # :. (du)/dx = x^3e^x + 3x^2e^x # The substituting into the integral [A], and applying the chain rule, we get: # d/dx int_0^(x^3e^x) \ (t^3+3)^17 \ dt = (du)/dx*d/(du) \ int_0^u \ (t^3+3)^17 \ dt # And now the derivative of the integral is in the correct form for the FTOC to be applied, giving: # d/dx int_0^(x^3e^x) \ (t^3+3)^17 \ dt = (du)/(dx)(u^3+3)^17# And restoring the initial substitution we get: # d/dx int_0^(x^3e^x) \ (t^3+3)^17 \ dt = (x^3e^x + 3x^2e^x)((x^3e^x)^3+3)^17# Which is option (2) in the initial question.

Charles Anctil · Answer

undefined The derivative of $ \int_{0}^{x^3e^x} (t^3+3)^{17} \, dt $ withTo find the derivative of the given integral with respect toThe derivative of $ \int_{0}^{x^3e^x} (t^3+3)^{17} \, dt $ with respect toTo find the derivative of the given integral with respect to $ x \The derivative of \( \int_{0}^{x^3e^x} (t^3+3)^{17} \, dt $ with respect to $To find the derivative of the given integral with respect to \( x $, we apply the Fundamental TheoremThe derivative of $ \int_{0}^{x^3e^x} (t^3+3)^{17} \, dt $ with respect to $ x $ is $ (x^3e^x)^3(e^x + 3)^{17} $.To find the derivative of the given integral with respect to $ x $, we apply the Fundamental Theorem of Calculus and the Chain Rule:

$$
\frac{d}{dx} \left( \int_0^{x^3e^x} (t^3 + 3)^{17} dt \right) = \frac{d}{dx} \left( F(x^3e^x) \right)
$$

where $ F(x) $ is the antiderivative of $ (t^3 + 3)^{17} $. Applying the Chain Rule:

$$
\frac{d}{dx} \left( F(x^3e^x) \right) = \frac{dF}{du} \cdot \frac{du}{dx}
$$

where $ u = x^3e^x $. Now, find $ \frac{dF}{du} $ and $ \frac{du}{dx} $:

$$
\frac{dF}{du} = (u^3 + 3)^{17} \cdot 3u^2 \cdot e^x
$$
$$
\frac{du}{dx} = (3x^2e^x + x^3e^x)
$$

Thus, the derivative is:

$$
\frac{d}{dx} \left( \int_0^{x^3e^x} (t^3 + 3)^{17} dt \right) = (x^3e^x)^3 + 3)^{17} \cdot 3(x^3e^x)^2 \cdot e^x \cdot (3x^2e^x + x^3e^x)
$$