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1. Integration by Parts We Have an Integration Method That “Undoes” the Chain Rule for Derivatives

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1. Integration by Parts We Have an Integration Method That “Undoes” the Chain Rule for Derivatives 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 − x2 ⋅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 ⋅ − π 2r 2xr x2dx = ∫0 − 2 1 3 2r = π ⋅ rx − 3 x S0 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.
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