Calculus: Analytic Geometry and Calculus, with Vectors

(lu) #1

(^208) Integrals
5 Look at the integral
f(1 + 5x)dx
and tell what must be done to enable us to evaluate the integral by
formula involving u".
6 Evaluate the integrals
(a) f (1 - x)3 dx (b) f sin 2x dx (c) f cos 3x dx
(d) f (1 - 2x) 2 dx (e) f e2z dx (f) f2x + 3dx
14ns..
use of the
(a) -(1 - x)4/4 + c (b) - cos 2x + c
(c) sin 3x + c (d) -g(1 - 2x)3 + c
(e) Te2z + c (f) y log I2x + 31 + c
7 Sometimes we can make small alterations in the way integrands are written
to put the integrands into forms where basic formulas are easily applied. Pay
careful attention to the examples
n
f tan x dx = fcos xdx -fcos x (^1) (-sin ) dx log cos xj + c
fx log xdx = f logx xdx = log Ilog xI +c
f xez' dx= f ez'(2x)dx = eye +
Then evaluate


(b) f

o z x

(a) f x fl -+ x2 dx dx

2

(^8) While the terminology plays a minor role in elementary calculus, we can
start learning that the equation
(1) dy= 2x
dx
is an example of an equation that should be called a derivative equation but is
called a differential equation. Functions y for which (1) holds are called solutions
of (1), and we know that (1) has many solutions. The particular solution of
(1) satisfying the boundary condition
(2) y = 16 when x = 3
is found in a straightforward way. If (1) holds, then
(3) y = f2x dx = x2 + C,
where c is a constant that can be 5 or- 3 or 416 but cannot be all of these things
at once. The function in (3) will satisfy (2) if 16= 9 + c and hence if c = 7.
Thus the answer is y = x2 + 7. With clues to methods being provided by this

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