Any assistance on calculating: the enclosed area $(A)$ between the curve $y=ln(x)$ and the $x-axis$ between $x = 1$ and $x = 2$ then to draw the graph of the function?
I know the equation will be something along the lines of:
$$A = \int_1^2{|ln(x)-0|} = \int_1^2{ln(x)}$$
Then:
$$\large\int{ln(x)} = xln(x)-x+C$$
$$\large A = (xln(x)-x)|_1^2 = 2(ln(2))-2-[1(ln(1))-1] = ?$$
But somehow my calculations seem incorrect and I am not sure how to proceed into graphing the function.
Best Answer
You went wrong when you changed base
$$\int_1^2ln(x)dx=\int_1^21\cdot ln(x)dx$$ Integration by parts leads you to
$$x\cdot ln(x)-\int x\cdot \frac{1}{x}dx$$ $$x\cdot ln(x)-x$$
plug in the x values to get an answer of 0.386
https://www.wolframalpha.com/input/?i=integrate+ln+x+from+1+to+2