How do you use the second fundamental theorem of Calculus to find the derivative of given #int [(ln(t)^(2))/t]dt# from #[3,x]#?

Answer 1

#int_3^xln^2(t)/tdt=(ln^3(x)-ln^3(3))/3#

First, note that using the substitution #u = ln(t) => du = 1/tdt# we have
#intln^2(t)/tdt = intu^2du = u^3/3+C = ln^3(t)/3+C#
The second fundamental theorem of calculus states that #F'(x) = f(x)=> int_a^bf(x)dx = F(b)-F(a)#
By the above, we have #d/dt ln^3(t)/3 = ln^2(t)/t#, and so
#int_3^xln^2(t)/tdt = ln^3(x)/3-ln^3(3)/3#
#=(ln^3(x)-ln^3(3))/3#
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Answer 2

Instead of actually finding the function and then differentiating, one could reason as follows:

The Second Fundamental Theorem of Calculus says that

If #f# is continuous on #[a,b]#, then
#int_a^b f(x) dx = F(b)-F(a)# where #F# is a function for which #F'(x) = f(x)# for all #x# in #[a,b]#.
In this case we are using the variable #t# in the integrand and the variable #x# as the upper limit of integration.
We want the derivative of #int_3^x [ln(t)^2/t] dt#.
Note that since we are asked about the interval #[3,x]#, we must have #x > 3# (otherwise the interval is either empty or undefined).
So, #ln(t)^2/t# is continuous on #[3,x]#, and
#int_3^x [ln(t)^2/t] dt= F(x) - F(3)# where #F# is a function such that #F'(x) = ln(x)^2/x#.
And there is our answer. The derivative of #int_3^x [ln(t)^2/t] dt# is #ln(x)^2/x#.

NOTE

The First Fundamental Theorem of Calculus says this directly. It says:

If #f# is continuous on #[a,b]# and #g# is defined for all #x# in #[a,b]# by,
#g(x) = int_a^x f(t) dt#, then #g# is continuous on #[a,b]# and #g'(x) = f(x)# for all #x# in #(a,b)#
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Answer 3

To find the derivative of the integral ( \int_{3}^{x} \frac{\ln(t)^2}{t} , dt ), you use the second fundamental theorem of calculus, which states that if ( f(x) ) is continuous on ([a, b]) and ( F(x) ) is an antiderivative of ( f(x) ) on ([a, b]), then:

[ \frac{d}{dx} \left( \int_{a}^{x} f(t) , dt \right) = f(x) ]

First, find an antiderivative ( F(x) ) of ( \frac{\ln(t)^2}{t} ). Let ( u = \ln(t) ), then ( du = \frac{1}{t} dt ). So, ( \int u^2 , du = \frac{u^3}{3} + C = \frac{\ln(t)^3}{3} + C ).

Therefore, ( F(x) = \frac{\ln(x)^3}{3} + C ).

Now, apply the second fundamental theorem of calculus:

[ \frac{d}{dx} \left( \int_{3}^{x} \frac{\ln(t)^2}{t} , dt \right) = \frac{d}{dx} \left( F(x) - F(3) \right) = \frac{d}{dx} \left( \frac{\ln(x)^3}{3} - \frac{\ln(3)^3}{3} \right) ]

[ = \frac{\frac{3\ln(x)^2}{x}}{3} = \frac{\ln(x)^2}{x} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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