Integration is a Calculus topic, and therefore, should not be posted in the Precalculus section. Anyway, to integrate ln(x), we use Integration by Parts (have you covered Integration by Parts yet?), i.e: [tex]\int u dv = uv - \int v du[/tex] We often use Integration by Parts, when no other methods can solve the integral. So, we want to integrate this: [tex]\int \ln (x) dx[/tex] We then let u = ln(x), and dv = dx So that implies du = dx / x, and v = x. Substitute all those into the formula, we have: [tex]\int \ln (x) dx = x \ln (x) - \int x \times \frac{dx}{x} = ...[/tex] Can you go from here? :)
lol i didnt learn those yet. BUt im glad it more complicated than it look. NO i cant do it. BUt i really appriciated the help. Instead i will try use simpsons rule. the question is area under the curve. 2day is monday-im at school school today so i quickly ask one of the mathematics teacher. =) i think he also hesitated say u can solve them by harder way but instead he told me simpson rule
It can be done without integration by parts. Try integrating over y instead of x, that is consider your graph to be x=exp(y).
This site is great for checking your work when integrating: http://integrals.wolfram.com It will not give you fully worked out answers but it will give you the final answer. Of course, your teacher will want you to solve problems showing your work all the way through, but this is still a great site.