(a) If <em>f(x)</em> is to be a proper density function, then its integral over the given support must evaulate to 1:

For the integral, substitute <em>u</em> = <em>x</em> ² and d<em>u</em> = 2<em>x</em> d<em>x</em>. Then as <em>x</em> → 0, <em>u</em> → 0; as <em>x</em> → ∞, <em>u</em> → ∞:

which reduces to
<em>c</em> / 2 (0 + 1) = 1 → <em>c</em> = 2
(b) Find the probability P(1 < <em>X </em>< 3) by integrating the density function over [1, 3] (I'll omit the steps because it's the same process as in (a)):

Answer:
17
Step-by-step explanation:
a[2]+b[2]=c[2]
8*8+15*15=hypotenuse.
64+225=289

hypotenuse=17
Answer:
0.012
Step-by-step explanation:
0.004 * 3 = 0.012
a = 4, b = -21, c = -18
to keep from getting "mixed up", evaluate the discriminant first ...
b<sup>2</sup> - 4ac = (-21)<sup>2</sup> - 4(4)(-18) = 729
sqrt(729) = 27
x = (21 +/- 27)/8
x = -3/4, x = 6
since the discriminant is a perfect square, the original quadratic will factor ...
4x<sup>2</sup> - 21x - 18 = 0
(4x + 3)(x - 6) = 0
x = -3/4, x = 6