How do you use the first part of the Fundamental Theorem of Calculus to find the derivative of #y = int 3(sin(t))^4 dt# from #e^x# to 1?

Question

Paisley Johnson · Accepted Answer

The derivative is #-3e^x(sin(e^x))^4#.

#y = int_(e^x)^(1)3(sin(t))^4dt# Let's let #f(t) = 3(sin(t))^4#. Recall that the antiderivative of a function can be used to calculate the definite integral of that function according to the calculus fundamental theorem. #int_a^b f(t) dt = F(b) - F(a)# where #F# is an antiderivative of #f#. Likewise: #int_(e^x)^1 f(t)dt = F(1) - F(e^x)# where #F# is an antiderivative of #f#. #y = F(1) - F(e^x)# We take the derivative because the question requests it. #color(red)(d/dx)y = color(red)(d/dx)(F(1)) - color(red)(d/dx)(F(e^x))# Don't be scared. #F(1)# is just a constant. #dy/dx = 0 - F'(e^x)*d/dx(e^x)# But by definition of an antiderivative, #F'(x) = f(x)#. #dy/dx = -e^xf(e^x)# #dy/dx = -e^x(3(sin(e^x))^4)# #dy/dx = -3e^x(sin(e^x))^4#

Cameron Alexander · Answer

undefined To find the derivative of $ y = \int_{e^x}^1 3(\sin(t))^4 \, dt $ using the first part of the Fundamental Theorem of Calculus, we first need to express the integral as a function of $ x $. We'll let $ F(x) $ be the antiderivative of $ 3(\sin(t))^4 $ with respect to $ t $. Then, by the Fundamental Theorem of Calculus, the integral $ \int_{e^x}^1 3(\sin(t))^4 \, dt $ is equal to $ F(1) - F(e^x) $.

To find $ F(x) $, we need to find the antiderivative of $ 3(\sin(t))^4 $ with respect to $ t $. This involves using trigonometric identities and integration techniques. After finding $ F(x) $, we evaluate $ F(1) $ and $ F(e^x) $, and then take the difference.

Finally, to find the derivative of $ y $ with respect to $ x $, we differentiate $ F(1) - F(e^x) $ using the chain rule, which states that if $ F(x) $ is differentiable at $ x $ and $ g(x) $ is differentiable at $ x $, then $ F(g(x)) $ is differentiable at $ x $ and $ (F(g(x)))' = F'(g(x)) \cdot g'(x) $. We compute $ \frac{d}{dx}[F(1) - F(e^x)] $ accordingly.