Given that the antiderivative of f of x equals 1 divided by x is F(x) = Ln|x| + C, evaluate the integral from 1 to 2 of the 1 divided by x, dx.

ln3
1/2
-1/2
ln2

∫1/x dx from 1 to 2

= lnx | from 1 to 2
= ln2 - ln1
= ln2 , since ln1 = 0

Q: Given that the antiderivative of f of x equals 1 divided by x is F(x) = Ln|x| + C, evaluate the integral from 1 to 2 of the 1 divided by x, dx?

A: Alrighty, let's crunch some numbers and tickle those integrals! To evaluate the integral from 1 to 2 of 1 divided by x, we can use the antiderivative formula. So, plugging in our values into the antiderivative, we get F(2) - F(1). Substituting 2 into the antiderivative formula gives us ln|2|, and substituting 1 gives us ln|1|.

Now, here comes the fun part: ln|2|. You see, the beauty of logarithms is that they're like little comedians, always hiding their true value. So the natural logarithm of 2 can be written as ln(2) or roughly 0.693.

So, F(2) - F(1) = ln|2| - ln|1| = ln(2) - ln(1) = ln(2) - 0 = ln(2).

Therefore, the integral from 1 to 2 of 1 divided by x, dx is equal to ln(2). So the answer is ln2! Enjoy the logarithmic hilarity!

To evaluate the integral from 1 to 2 of 1/x, we can use the antiderivative formula.

The antiderivative of 1/x is ln|x| + C.

Using the Fundamental Theorem of Calculus, we can evaluate the definite integral by subtracting the value of the antiderivative at the lower bound from the value at the upper bound.

So, evaluating the integral from 1 to 2 of 1/x, we get:

F(2) - F(1) = ln|2| - ln|1|

Since ln|1| = 0, we are left with:

ln|2|

So, the value of the integral from 1 to 2 of 1/x is ln|2|.

To evaluate the integral from 1 to 2 of 1/x, we can use the fundamental theorem of calculus and subtract the values of the antiderivative at the upper and lower limits.

Using the antiderivative F(x) = Ln|x| + C, we have:

∫ (1/x) dx = F(x) + C

∫ (1/x) dx = Ln|x| + C

To evaluate the definite integral from 1 to 2, we substitute 2 and 1 into the antiderivative and subtract the values:

∫[1,2] (1/x) dx = [Ln|x|]₂ - [Ln|x|]₁

Substituting 2 into the antiderivative:

[Ln|2|]₂

Since the natural logarithm of a positive number is defined as the logarithm of the number itself, we have:

[Ln(2)]₂

Now substituting 1 into the antiderivative:

- [Ln|1|]₁

Since the natural logarithm of 1 is 0, we have:

- (0)

Therefore, the evaluated integral from 1 to 2 of 1/x is:

[Ln(2)]₂ - (0) = Ln(2)

So, the answer is ln(2).