Finding the area under xf(x) for a piecewise function

[karush]

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]

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]

Since f(x)= x if $$x\le 0$$ and f(x)= x+ 1 if x> 0, $$xf(x)= x^2$$ if $$x\le 0$$ and $$xf(x)= x^2+ x$$ if x> 0. So $$\int_{-2}^1 xf(x)dx= \int_{-2}^0 x^2 dx+ \int_0^1 x^2+ x dx$$.

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

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

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

 

[karush]

I thot of doing that but didn't think it was correct

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

 

[karush]

Since f(x)= x if $$x\le 0$$ and f(x)= x+ 1 if x> 0, $$xf(x)= x^2$$ if $$x\le 0$$ and $$xf(x)= x^2+ x$$ if x> 0. So $$\int_{-2}^1 xf(x)dx= \int_{-2}^0 x^2 dx+ \int_0^1 x^2+ x dx$$.

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

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

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

mahalo

 

[tkhunny]

I thot of doing that but didn't think it was correct

These exam questions you 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.