1. Integration by parts We have an integration method that “undoes” the chain rule for derivatives. We now present a method that “undoes” the product rule for derivatives. The method is know as integration by parts.

From the product rule, we know that

′ ′ ′ [f(x) ⋅ g(x)] = f(x)g (x) + g(x)f (x).

Rearranging, we get

′ ′ ′ f(x)g (x) = [f(x) ⋅ g(x)] − g(x)f (x).

Now we integrate both sides to get

′ ′ S f(x) ⋅ g (x)dx = f(x) ⋅ g(x) − S g(x) ⋅ f (x)dx.

du ′ dv ′ Finally, call u = f(x) and v = g(x). Then dx = f (x) and dx = g (x). Then we get the very important formula

S u ⋅ dv = u ⋅ v − S v ⋅ du.

Example 1. Find ∫ ln(x)dx.

This may not look like a formula of the form u ⋅ dv, but it is. If we let u = ln(x) 1 and v = x, then du = x dx and dv = 1dx. So

S ln(x)dx = S u ⋅ dv = u ⋅ v − S v ⋅ du = ln(x) ⋅ x − S 1dx = ln(x) ⋅ x − x + C.

To reiterate, ∫ ln(x)dx = x ln(x) − x + C.

Example 2. Find ∫ x cos(x)dx.

The challenge of integration by parts problems is to determine which function should be u and which should be dv. Let’s see what happens when we let u = cos(x) 1 2 and dv = xdx. Then we get v = 2 x and du = −sin(x)dx. Then

1 2 1 1 x cos x dx x2 cos x x2 sin x dx. S ( ) = 2 ( ) − S −2 ( ) While this formula is true, our choices of u and dv yielded a much more com- plicated formula. So let’s switch them. u = x and dv = cos(x)dx. So du = 1dx and v = sin(x). So we get

S x cos(x)dx = x sin(x) − S sin(x)dx = x sin(x) + cos(x) + C.

As with all antiderivative problems, we can check our answer with differentiation.

The integration by parts formula is worth memorizing. Some people use the mneumonic “ultra-violet voodoo” to help remembering it.

2. Volume of a sphere We are about to take our first steps into 3-dimensional geometry. Just as the circle is a fundamental object in 2 dimensions, its analog, the sphere, is just as fundamental to 3-dimensional study. Remember that we thought of area as a func- tion from the plane to the real numbers. We have an analog in three dimensions. We will think of volume as a function from a 3-dimensional “plane” to the real numbers. Let’s begin by determining the volume of a sphere.

Suppose our sphere will have radius r. We begin by imagining the upper half of the circle of radius r centered at r, 0 . The equation for this curve is x r 2 y r2 √ ( ) ( − ) + = or y = −x2 + 2xr. See figure 5.33 in your text for reference. Let’s pick some x ∈ [0, 2r], and let dx be some small horizontal value. Consider the rectangle determined by (x, y) and dx, that is, the rectangle of height y and width dx. The area of this rectangle is y ⋅ dx. Now imagine we take this rectangle and swing it around the x-axis, creating a short cylinder skewered by the x-axis. See figure 1 below for visualization. √ The radius of the base of this cylinder is y = 2xr − x2, and the height is dx. So √ 2 the volume of this cylinder is π ⋅‰ 2xr − x2Ž ⋅dx. Finally, imagine that we do this for all x and let dx approach zero. Then, the volume of the sphere with radius r is given by

√ 2 2r π 2xr x2 dx ∫0 ⋅ ‰ − Ž π 2r 2xr x2dx = ∫0 −

2 1 3 2r = π ⋅ ‰rx − 3 x S0 Ž 3

3 1 3 = π ⋅ (4r − 3 8r )

4 3 = 3 πr

Figure 1

4 3 So we have shown that the volume of a sphere of radius r is 3 π ⋅ r , as we expected.

3. Volumes of revolution Suppose that f(x) is a function on [a, b] with f(x) ≥ 0. Let x ∈ [a, b] and let y = f(x). Imagine we have a rectangle of width dx and height y = f(x). Then the area of the rectangle will be f(x) ⋅ dx. As before, imagine swinging this rectangle around the x-axis to get a thin cylinder skewered by the x-axis. The radius of the base of the cylinder will be f(x), and the height is dx. So the volume is given by π ⋅ [f(x)]2 ⋅ dx. See figure 1 again for visualization.

Now we again do this for all x and let dx get arbitrarily small and add up the volumes of the resulting disks. The volume of the resulting 3-dimensional solid will be given by

b 2 V = π [f(x)] dx. Sa 4 √ Example 3. Find the volume of the solid given by the curve f(x) = x on [0, 4] revolved around the x-axis.

4 4 √ 2 1 2 4 1 The volume is given by V = π [ x] dx = π xdx = π ‹ x S0 = π⋅ ⋅16 = 8π. S0 S0 2 2

Example 4. Find the volume of the solid given by the curve f(x) = sin(x) on [0, π] revolved around the x-axis.

π 2 V = π sin (x)dx. We need a trigonometric identity, namely a half angle S0 π 1−cos(2x) 1 1 cos(2x) 2 formula, sin(x) = 2 = 2 − 2 − 2 . Using this we get, π sin (x) = S0 π 1 cos(2x) 1 1 π π π π − dx = π ‹ x − sin(2x)S0  = − 0 − (0 − 0) = . S0 2 2 2 4 2 2