Calculus: Analytic Geometry and Calculus, with Vectors

9.5 Integration by parts 525

10 Evaluate the integral in

f

x3

I J

1+xsdx

in two different ways, and make the results agree. First, use the identity

x3 x(l + x2 - 1) x

1+x2

-x 1+x2-

1+x2 1+x2

Second, integrate by parts with

u = x2, de = x(1 + x2)-1,J dx.

11 With the aid of the substitution (or change of variable) = t,

show that

fa

x2 dx = 2 '(a - t2)2t2 dt = Trga.

o fo

Show that the first integral can be evaluated by integration by parts.

12 Derive the formulas

f

ine-0/2 dt = x lez'12 - I

t-2e :'12 dt

x x

f e-"/2 dt = xlex212 - x3ex'12 + 3 t-4e 1=1'- dt.

x x

Remark: Formulas of this nature give useful information, the integrals on the

right being small in comparison to those on the left when x exceeds 2 or 3 or 4.

13 Many problems in pure and applied mathematics involve "best fit" or

"best approximation" in some sense or other. We can start picking up ideas by

observing that if X is near 1 and

(1) f(x) =cosx,g(x) =(1 - a2)

then the graphs of f(x) and Xg(x) over the interval -a/2 < x < 7r/2 look much

alike. The graph of Xg(x) is the best fit in the sense of least squares to the graph of

f (x) when X is chosen such that

(2) f Ax/2 [f(x) - Xg(x)]2 dx

is a minimum. Show that (2) will be a minimum when

(3)

a

row/2

[g(x)]2 dx =f07/2f(x)g(x) dx

and hence when X =310/?r3= 0.967546. Remark: Sketching graphs of f (x)

and g(x) on a rather large scale indicates that g(x) > f(x) when 0 < x < 7r/2.

In fact, we can put

(4) F(x) = g(x) - f (x)

and show that F(0) = F(ir/2) = 0, F is increasing over the part of the interval

0 < x < r/2 for which sin x > (8/ir2)x, and F is decreasing over the remaining

part of the interval.