How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #y = int sin^3 t dt# from #[e^x, 0]#?

Question

Isabella Ackerman · Accepted Answer

The theorem says: Given the function: #y=int_(h(x))^g(x)f(t)dt# than: #y'=f(g(x))*g'(x)-f(h(x))*h'(x)#. So, since #y=int_(e^x)^0sin^3tdt#: #y'=sin^3 0*0-sin^3e^x*e^x=-e^xsin^3e^x#.

Emilia Aguilar · Answer

undefined To use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function $ y = \int_{e^x}^{0} \sin^3(t) \, dt $, we first need to evaluate the integral and then differentiate the result with respect to $ x $. Let's denote the integral as a function of $ x $, say $ F(x) $. Then, according to the Fundamental Theorem of Calculus Part 1, the derivative of $ F(x) $ with respect to $ x $ is given by $ \frac{d}{dx} \int_{e^x}^{0} \sin^3(t) \, dt = -\sin^3(e^x) \cdot \frac{d}{dx}(e^x) $.

Differentiating $ e^x $ with respect to $ x $ gives $ \frac{d}{dx}(e^x) = e^x $.

So, the derivative of the function $ y = \int_{e^x}^{0} \sin^3(t) \, dt $ with respect to $ x $ is $ -\sin^3(e^x) \cdot e^x $.