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[SOLVED] 4.5.012 AP Calculus Exam ... area of piece wise funcion

karush

Well-known member
Jan 31, 2012
2,678
View attachment 9552
ok well it isn't just adding the areas of 2 functions but is $xf(x)$ as an integrand

Yahoo had an answer to this but its never in Latex so I couldn't understand how they got $\dfrac{7}{2}$
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
ok well it isn't just adding the areas of 2 functions but is $xf(x)$ as an integrand

Yahoo had an answer to this but its never in Latex so I couldn't understand how they got $\dfrac{7}{2}$
What have you tried? Why can't you split it up at x = 0. Is xf(x) ever negative on that interval?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Since f(x)= x if [tex]x\le 0[/tex] and f(x)= x+ 1 if x> 0, [tex]xf(x)= x^2[/tex] if [tex]x\le 0[/tex] and [tex]xf(x)= x^2+ x[/tex] if x> 0. So [tex]\int_{-2}^1 xf(x)dx= \int_{-2}^0 x^2 dx+ \int_0^1 x^2+ x dx[/tex].

[tex]\int_{-2}^0 x^2 dx= \left[\frac{1}{3}x^3\right]_{-2}^0= \frac{1}{3}(0- (-8))= \frac{8}{3}[/tex].

[tex]\int_0^1 x^2+ x dx= \left[\frac{1}{3}x^3+ \frac{1}{2}x^2\right]_0^1= \frac{1}{3}+ \frac{1}{2}= \frac{5}{6}[/tex].

[tex]\int_{-2}^1 xf(x) dx= \frac{8}{3}+ \frac{5}{6}= \frac{16}{6}+ \frac{5}{6}= \frac{21}{6}= \frac{7}{2}[/tex].
 

karush

Well-known member
Jan 31, 2012
2,678
I thot of doing that but didn't think it was correct🤔

These exam questions ya got to be very careful they are very subtle
 

karush

Well-known member
Jan 31, 2012
2,678
Since f(x)= x if [tex]x\le 0[/tex] and f(x)= x+ 1 if x> 0, [tex]xf(x)= x^2[/tex] if [tex]x\le 0[/tex] and [tex]xf(x)= x^2+ x[/tex] if x> 0. So [tex]\int_{-2}^1 xf(x)dx= \int_{-2}^0 x^2 dx+ \int_0^1 x^2+ x dx[/tex].

[tex]\int_{-2}^0 x^2 dx= \left[\frac{1}{3}x^3\right]_{-2}^0= \frac{1}{3}(0- (-8))= \frac{8}{3}[/tex].

[tex]\int_0^1 x^2+ x dx= \left[\frac{1}{3}x^3+ \frac{1}{2}x^2\right]_0^1= \frac{1}{3}+ \frac{1}{2}= \frac{5}{6}[/tex].

[tex]\int_{-2}^1 xf(x) dx= \frac{8}{3}+ \frac{5}{6}= \frac{16}{6}+ \frac{5}{6}= \frac{21}{6}= \frac{7}{2}[/tex].
mahalo
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
I thot of doing that but didn't think it was correct🤔

These exam questions ya got to be very careful they are very subtle
This one is not anywhere near "subtle". It wants to know if you know that you can partition the Domain of the integral. This is a very important, fundamental principle.