How do you use the second fundamental theorem of Calculus to find the derivative of given #int ((sin^3)(t))dt# from #[0, e^x]#?

Question

Luke Alvarez · Accepted Answer

#e^x sin^3 e^x #

you want

#d/dx int_0^(e^x) sin^3(t) \ dt#

by the FTC we know that

#d/(du) int_a^u f(t) \ dt = f(u)#

and if #u = u(x)# we also know that

#d/(dx) int_a^u f(t) \ dt = f(u)* (du)/dx#

we apply that here with #f(t) = sin^3 t#, #a = 0# and #u = e^x# to get

#sin^3 e^x * e^x#

Ezekiel Alexander · Answer

undefined To use the Second Fundamental Theorem of Calculus to find the derivative of $ \int_{0}^{e^x} \sin^3(t) \, dt $, first, you need to find the antiderivative of the integrand, $ \sin^3(t) $. Then, apply the Second Fundamental Theorem of Calculus, which states that if $ f $ is continuous on $[a, b]$ and $ F $ is an antiderivative of $ f $ on $[a, b]$, then $ \int_{a}^{b} f(x) \, dx = F(b) - F(a) $.

The antiderivative of $ \sin^3(t) $ can be found by using the power reduction formula for trigonometric functions. The power reduction formula for $ \sin^3(t) $ is $ \sin^3(t) = \frac{3\sin(t) - \sin(3t)}{4} $.

Now, integrate this expression with respect to $ t $ from $ 0 $ to $ e^x $:

$$ \int_{0}^{e^x} \frac{3\sin(t) - \sin(3t)}{4} \, dt $$

After integrating and evaluating at the limits of integration, you'll obtain the derivative of the given integral with respect to $ x $.

Emma Aldridge · Answer

undefined To find the derivative of the integral $ \int_{0}^{e^x} \sin^3(t) \, dt $ with respect to $ x $, we can apply the second fundamental theorem of calculus. According to this theorem, if $ F(x) $ is a continuous function on the interval $[a, b]$, and $ f(x) $ is a continuous function on an interval containing $[a, b]$, then:

$$ \frac{d}{dx} \left( \int_{a}^{u(x)} f(t) \, dt \right) = f(u(x)) \cdot \frac{du}{dx} $$

Given that $ u(x) = e^x $ and $ f(t) = \sin^3(t) $, we can proceed as follows:

$$ \frac{d}{dx} \left( \int_{0}^{e^x} \sin^3(t) \, dt \right) = \sin^3(e^x) \cdot \frac{d}{dx}(e^x) $$

$$ = \sin^3(e^x) \cdot e^x $$

Therefore, the derivative of the given integral $ \int_{0}^{e^x} \sin^3(t) \, dt $ with respect to $ x $ is $ \sin^3(e^x) \cdot e^x $.