Jan 3, 2020 Thread starter #1 karush Well-known member Jan 31, 2012 2,678 ok I got stuck real soon..... .a find where the functions meet $$\ln x = 5-x$$ e both sides $$x=e^{5-x}$$ ok how do you isolate x? W|A returned $x \approx 3.69344135896065...$ but not sure how they got it b.? c.? Last edited: Jan 3, 2020
ok I got stuck real soon..... .a find where the functions meet $$\ln x = 5-x$$ e both sides $$x=e^{5-x}$$ ok how do you isolate x? W|A returned $x \approx 3.69344135896065...$ but not sure how they got it b.? c.?
Jan 3, 2020 #2 skeeter Well-known member MHB Math Helper Mar 1, 2012 655 FYI, this is a calculator active problem ... $y = \ln{x} \implies x = e^y$ $y = 5-x \implies x = 5-y$ \(\displaystyle R = \int_0^{1.3065586} (5-y) - e^y \, dy = 2.986\) Attachments regionR.png 1.6 KB Views: 14 AreaR.png 1.8 KB Views: 15
FYI, this is a calculator active problem ... $y = \ln{x} \implies x = e^y$ $y = 5-x \implies x = 5-y$ \(\displaystyle R = \int_0^{1.3065586} (5-y) - e^y \, dy = 2.986\)
Jan 4, 2020 Thread starter #3 karush Well-known member Jan 31, 2012 2,678 so then we don't need to know x of the intersection
Jan 4, 2020 #4 skeeter Well-known member MHB Math Helper Mar 1, 2012 655 karush said: so then we don't need to know x of the intersection Click to expand... yes, you need both x and y coordinates of the intersection point ... (a) can be done w/r to x or y ... I went w/r to y because it only requires a single integral expression (b) requires x ... two integrals (c) requires y
karush said: so then we don't need to know x of the intersection Click to expand... yes, you need both x and y coordinates of the intersection point ... (a) can be done w/r to x or y ... I went w/r to y because it only requires a single integral expression (b) requires x ... two integrals (c) requires y
Jan 8, 2020 #6 skeeter Well-known member MHB Math Helper Mar 1, 2012 655 karush said: ill go with c Click to expand... ? (c) is a free response question, not a choice