THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS
157
20.47
Calculate
By Problem 20.44, the derivative is
20.48 If/is an odd function, show that
Let u = — x, du = —dx.
Hence,
Then
20.49 Evaluate
x
2 sin jc is an odd function, since sin (—x) = —sin x. So, by Problem 20.48, the integral is 0.
20.50 If/is an even function, show that
Let u = — x. du = —dx.
Hence,
Then
20.51 Find
By Problem 20.44, the derivative is
20.52 Solve
for b.
Hence, 2 = b"-l, 6" =3, 6 =
20.53
If
compute
Then
20.54
If
find
for
for
20.55 Given that
find a formula tor f(x) and evaluate a.
First, set x = a to obtain 2o
2 -8 = 0, a
2 = 4, a = ±2. Then, differentiating, we find 4x = /(.x).
20.56
Given H(x) =
find //(I) and H'(\), and show that //(4) - H(2) < I.
for some c in (2,4). Now, H'(c) =
By the Mean-Value Theorem, H(4) - H(2) =
Hence, tf(4) - H(2) < i.
20.57
If the average value of f(x) = x
3 + bx - 2 on [0,2] is 4, find b.
(jt
3 + bx - 2) dx = 4(i*
4 + ^6x
2 - 2x) ]
2 = H(4 + 26 - 4) - 0] = 6. Thus,
20.58 Find
Therefore, the desired limit is
Then
so, ff'U)=i/(^) x
2 sin x dx.
f(x)dx = 2 f(x) dx.
dx = 2/n
f(x-k)dx = l,
Let x = u — fc, rfx = dw.
/W«fa =
/(« -k)du = f(x-k)dx=l.
/W =
sinx
3x
2
x<0
x>0
/(x) dx.
f(x)dx =
(-0+1) + 1 = 2.
(4-2)-//'(c)
dt.
//'W =
rff = 0.
//(!) =
6 = 4
Let g(jc) =
4 =