8.4. EXERCISES 197

Example 8.3.8 Suppose the velocity is v(t) = t − t3. Find the distance the object moves onthe real line for t ∈ [0,2].

As just explained, it is∫ 2

0

∣∣t − t3∣∣dt. You must split this up into intervals on which you

can remove the absolute values. t−t3 ≥ 0 on [0,1] and it is ≤ 0 on [1,2] so the total distancetravelled is ∫ 1

0

(t − t3)dt +

∫ 2

1

(−t + t3)dt =

52

Sometimes people want to use a shortcut on problems like this. They want to say that anantiderivative is

∣∣∣ t2

2 − t4

4

∣∣∣ and then plug in the end points and evaluate. This is totally wrong

because the function just described is not an antiderivative of the function t →∣∣t − t3

∣∣!8.4 Exercises1. Find the following antiderivatives.

(a)∫

x3e−3x dx(b)

∫x4 cosxdx

(c)∫

x5ex dx

(d)∫

x6 sin(2x) dx

(e)∫

x3 cos(x2)

dx

2. Find the following antiderivatives.

(a)∫

xe−3x dx

(b)∫ 1

x(ln(|x|))2 dx

(c)∫

x√

2− xdx

(d)∫(ln |x|)2 dx Hint: Let u(x) = (ln |x|)2 and v′ (x) = 1.

(e)∫

x3 cos(x2)

dx

3. Show that∫

sec3 (x) dx =

12

tan(x)sec(x)+12

ln |secx+ tanx|+C.

4. Find∫ xex

(1+x)2 dx.

5. Consider the following argument. Integrate by parts, letting u(x) = x and v′ (x) = 1x2

to get ∫ 1x

dx =∫

x(

1x2

)dx =

(−1

x

)x+

∫ 1x

dx =−1+∫ 1

xdx.

Now subtracting∫ 1

x dx from both sides, 0 = −1. Is there anything wrong here? Ifso, what?

6. Find the following antiderivatives.