Originally Posted by
evansf Given
first sketch the graph
y = f(x) on the given interval. Then find the integral of f using your knowledge of area formulas for
rectangles, triangles and circles.
$\displaystyle \int_{-2}^4 |3 - |3x||dx = \int_{-2}^4 |3 - 3|x||dx = \int_{-2}^4 3|1 - |x||dx$.. now, this can be separated to
$\displaystyle \int_{-2}^4 |3 - |3x||dx = ... = \int_{-2}^0 3|1 - |x||dx + \int_{0}^4 3|1 - |x||dx$..
we know, in the first addend, x is negative, hence we take |x| = -x; on the other hand, in the second addend, x is positive hence |x| = x
so the integral would be
$\displaystyle \int_{-2}^0 3|1 - |x||dx + \int_{0}^4 3|1 - |x||dx = \int_{-2}^0 3|1 + x|dx + \int_{0}^4 3|1 - x|dx$
now, we have to check where |1-x| is positive or negative and |1+x| is positive or negative..
for |1-x|, we take 1-x if 1-x >= 0 or x <= 1; we take x-1 otherwise
for |1+x|, we take 1+x if 1+x >= 0 or x >= -1; we take -x-1 otherwise
so now,
$\displaystyle ... = \int_{-2}^{-1} 3(-1 - x)dx + \int_{-1}^{0} 3(1 + x)dx + \int_{0}^1 3(1 - x)dx + \int_{1}^4 3(x - 1)dx$
so can you do it now??