Section 7.5 Review on Integrals
Ruipeng Shen
Jan 6
1
Basic Definition
Indefinite Integral The indefinite integral of f (x) gives its anti-derivatives.
F 0 (x) = f (x) in an open interval I, then we have
Z
f (x) dx = F (x) + C.
Namely, if
Definite Integral The definite integral of a bounded function f (x) on an interval [a, b] is
defined by the limit of the Riemann sums.
Z
b
f (x)dx = lim
n→∞
a
n
X
f (x?i )∆x.
i=1
If f (x) ≥ 0, this can also be understood as the area of the region bounded below the graph
y = f (x).
Δx=(b-a)/n
x0 x1* x1
2
xi=a+iΔx
Ai=f(x*i)·Δx
f(x*i)
x2
xi-1 x*i xi
xn-1 xn
The Fundamental Theorem of Calculus
If F 0 (x) = f (x) is a continuous function, then we have
Z
b
f (x) dx = F (b) − F (a).
a
1
3
Integration by Basic Formulas
Z
kdx = kx + C
Z
Z
xn dx =
xn+1
+ C (n 6= −1)
n+1
Z
1
dx = ln |x| + C
x
Z
ax
ax dx =
+C
ln a
Z
cos x dx = sin x + C
Z
csc2 x dx = − cot x + C
Z
csc x cot x dx = − csc x + C
Z
1
√
dx = arcsin x + C
1 − x2
Z
cosh x dx = sinh x + C
ex dx = ex + C
Z
sin x dx = − cos x + C
Z
sec2 x dx = tan x + C
Z
sec x tan x dx = sec x + C
Z
1
dx = arctan x + C
1 + x2
Z
sinh x dx = cosh x + C
Z
Example 1. Show that
0
Solution
4
b
π
dx
< for all b > 0.
2
1+x
2
By the fundamental theorem of calculus and the basic formula, we have
Z b
dx
π
b
= [arctan x]0 = arctan b < .
2
1
+
x
2
0
The u-substitution
If u = g(x), then we have
Z
Z
Example 2. Find
Solution
5
f (g(x))g 0 (x) dx =
Z
f (u) du
ex
dx
1 + ex
Let u = ex + 1, thus we have du = ex dx.
Z
Z
ex
du
dx =
= ln(ex + 1) + C.
1 + ex
u
Integration by Parts
Z
Z
udv = uv −
Z
Example 3. Evaluate
x tan2 x dx.
2
vdu.
Solution
Applying integration by parts, we have
Z
Z
Z
sin2 x
1
x tan2 x dx = x ·
dx
=
(x
sin
x)d
cos2 x
cos x
Z
x sin x
1
=
−
d(x sin x)
cos x
cos x
Z
x cos x + sin x
dx
=x tan x −
cos x
x2
=x tan x −
+ ln | cos x| + C
2
Alternatively we have
Z
Z
Z
Z
x2
x d(tan x) −
x tan2 x dx = x sec2 x dx − x dx =
2
Z
x2
=x tan x −
tan x dx −
2
2
x
=x tan x + ln | cos x| −
+C
2
6
Integration by Inverse Substitution
Z
Example 4. Evaluate
Solution
√
x3
dx.
1 + x2
√
Let x = tan θ, thus 1 + x2 = sec θ and dx = sec2 θdθ.
Z
Z
x3
tan3 θ
√
· sec2 θ dθ
dx =
sec θ
1 + x2
Z
= (sec2 θ − 1) · (sec θ tan θ dθ)
Z
p
= (u2 − 1) du
(u = sec θ = 1 + x2 )
u3
−u+C
3
p
1
= (1 + x2 )3/2 − 1 + x2 + C.
3
=
7
Integration by Partial Fractions
Z
Example 5. Evaluate the integral
4x
dx.
x3 + x2 + x + 1
Solution Since we can factor the denominator as x3 + x2 + x + 1 = (x + 1)(x2 + 1), we have
the partial fractional decomposition
x3
4x
A
Bx + C
=
+ 2
.
+ +x+1
x+1
x +1
x2
3
Multiplying both sides by (x + 1)(x2 + 1), we have
4x = A(x2 + 1) + (Bx + C)(x + 1)
Plugging in x = −1, we have A = −2. This implies
Bx + C =
4x − A(x2 + 1)
4x + 2(x2 + 1)
=
= 2x + 2.
x+1
x+1
Thus we have
Z
4x
dx =
3
2
x +x +x+1
Z −2 2x + 2
+ 2
x
x +1
dx
= − 2 ln |x| + ln(x2 + 1) + 2 arctan x + C.
4