this case, we simply replace t in the integrand with x: f′(x) = x^3 + x. Now, all we have to do is
plug in x = 5: f′(5) = 5^3 + 5 = 130.
42. C This is a simple integral that we do using integration by parts. You should memorize that ∫ ln
(ax) dx = x ln (ax) − x + C, which makes this integral easy.
Step 1: The formula for integration by parts is ∫ u dv = uv − ∫ v du.
The trick is that we have to let dv = dx.
Let u = ln 2x and dv = dx
du = dx = dx and v = x
Plugging in to the formula we get
∫ In 2x dx = x In 2x − ∫^ dx = x In 2x − x + C
- B First, in order to be differentiable, f has to be continuous. Here, this means that if we plug 2
into both pieces of f, we need to get the same value: a(2)^2 + 3b(2) + 14 = 3a(2) + 5b, which
simplifies to
4 a + 6b + 14 = 6a + 5b
2 a − b = 14
Next, in order to be differentiable, if we plug 2 into both pieces of f′, we need to get the same
value. First, differentiate both pieces of f : f′(x) = . Now we plug in 2
and set the two pieces equal to each other: 4a + 3b = 3a, so a = −3b. Plug this into the first
equation: 2(−3b) − b = 14. Thus, b = −2 and a = 6.
- D If you want to find acceleration, all you have to do is take the second derivative of the position
functions.
Step 1: = −2 sin 2t and = 4 cos t
= − 4 cos 2t and = −4 sin t
