How do you use the Fundamental Theorem of Calculus to find the derivative of #int 5(sin(t))^5 dt# from 5 to e^x?

Question

Christopher Allers · Accepted Answer

#= 5(sin(e^x))^5 * e^x#

we want #d/dx int_5^{e^x} 5(sin(t))^5 dt# using FTC

FTC tells us that #d/(du) int_a^u f(t) dt = f(u)#

here a = 5, u = e^x

so FTC + chain rules tells us that

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

#= 5(sin(u))^5 * e^x#

#= 5(sin(e^x))^5 * e^x#

Aurora Alvarado · Answer

#int_5^(e^x)5sin^5tdt=-5{cose^x-2/3cos^3e^x+1/5cos^5e^x}+5{cos5-2/3cos^3(5)+1/5cos^5(5)}.# Fundamental Theorem of Calculus : #intf(x)dx=F(x)+C rArr int_a^bf(x)dx=[F(x)]_a^b=F(b) -F(a)# So, let us first find #I=int5sin^5tdt=int5sin^4t*sintdt=int5(1-cos^2t)^2*sintdt............(1).# Now take substitution #cost=y,# so that #-sintdt=dy. Hence, #I=5int(1-y^2)^2*(-dy)=-5int(1-2y^2+y^4)dy=-5(y-2/3y^3+1/5y^5)+C =-5(cost-2/3cos^3t+1/5cos^5t)+C.# Finally, #int_5^(e^x)5sin^5tdt=-5{cose^x-2/3cos^3e^x+1/5cos^5e^x}+5{cos5-2/3cos^3(5)+1/5cos^5(5)}.#

Aurora Alvarado · Answer

undefined To find the derivative of the integral $ \int_{5}^{e^x} \sin(t)^5 \, dt $, you can use the Fundamental Theorem of Calculus. According to the theorem, if $ F(x) $ is any antiderivative of $ f(x) $, then $ \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) $. In this case, $ f(t) = \sin(t)^5 $.

First, find an antiderivative of $ f(t) $, which is $ F(t) $. Then evaluate $ F(e^x) - F(5) $. Finally, take the derivative of this expression with respect to $ x $.

Let's find $ F(t) $:
$$ F(t) = \int \sin(t)^5 \, dt $$

Now, we find $ F(e^x) - F(5) $:
$$ F(e^x) - F(5) = \left[ \int \sin(t)^5 \, dt \right]_{t=5}^{t=e^x} $$

Finally, take the derivative of this expression with respect to $ x $. This will give you the derivative of the given integral with respect to $ x $.