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

#### karush

##### Well-known member
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
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
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

##### Well-known member
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
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

##### Well-known member
MHB Math Helper
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.