Properties of Definite Integrals

1. The variable \(x\) that appears in definite integrals is called the dummy variable and we can replace this with another to get the same result:
\(
\int_a^b f(x) \mathrm{d} x=\int_a^b f(t) \mathrm{d} t .
\)

Proof:

\(
\begin{aligned}
L H S &=\int_a^b f(x)+c \\
&=[g(x)+c]_a^b \\
&=[g(b)+c]-[g(a)+c] \\
&=g(b)-g(a)
\end{aligned}
\) \(
\begin{aligned}
\text { RHS } &=\int_a^b f(t) d t \\
&=[g(t)+c]_a^b \\
&=[g(b)+c]-[g(a)+c] \\
&=g(b)-g(a)
\end{aligned}
\)

Thus LHS = RHS

2. The definite integral of a constant \(c \in \mathbf{R}\) is proportional to the width of the interval:
\(\int_a^b c \mathrm{~d} x=c(b-a)\)

Proof:

\(
\begin{aligned}
\int_a^b c d x \\
&=c [x]_a^b \\
&=c(b-a) \\
\end{aligned}
\)

3. The zero-width limit when \(b=a\) implies that
\(
\int_a^a f(x) \mathrm{d} x=0 .
\)

Example 1:

\(
\begin{aligned}
\int_2^2 2 x  d x &=\left[x^2\right]_2^2 \\
&=2^2-2^2 \\
&=0
\end{aligned}
\)

4. We can interchange the limits on any definite integral using
\(
\int_a^b f(x) \mathrm{d} x=-\int_b^a f(x) \mathrm{d} x .
\)

Proof:

\(
\begin{aligned}
L H S &=\int_a^b f(x)+c \\
&=[g(x)+c]_a^b \\
&=[g(b)+c]-[g(a)+c] \\
&=g(b)-g(a)
\end{aligned}
\) \(
\begin{aligned}
L H S &=-\int_b^a f(x)+c \\
&=-[g(x)+c]_b^a \\
&=-([g(a)+c]-[g(b)+c]) \\
&=g(b)-g(a)
\end{aligned}
\)

LHS = RHS

5. We can split up definite integrals with a sum or difference:
\(
\int_a^b(f(x) \pm g(x)) \mathrm{d} x=\int_a^b f(x) \mathrm{d} x \pm \int_a^b g(x) \mathrm{d} x
\)

Example 2: Integral of Sum of functions: Evaluate the integral \(\int\left[x+e^x\right] d x\)

According to the above property

\(\int_a^b(f(x) + g(x)) \mathrm{d} x=\int_a^b f(x) \mathrm{d} x + \int_a^b g(x) \mathrm{d} x \\\) \(
\int\left[x+e^x\right] d x=\int x d x+\int e^x d x \\
=x^2 / 2+e^x+c 
\)

 

Proof: Integral of difference of functions

Evaluate the integral
\(
\int[2-1 / x] d x
\)
\(
\int[2-1 / x] d x=\int 2 d x-\int(1 / x) d x
\)
\(
=2 x-\ln |x|+c
\)

6. We can factor out a constant \(c \in \mathbf{R}\) from definite integrals:
\(\int_a^b c f(x) \mathrm{d} x=c \int_a^b f(x) \mathrm{d} x .\)

Example 3:

\(\int_1^2 5 f(x) \mathrm{d} x=5 \int_1^2 f(x) \mathrm{d} x . \)

 

7. We can also split up the integral with the limits \([a, b]\) for some value \(c \in \mathbf{R}\) over two adjacent intervals \([a, c]\) and \([c, b]\):

\(
\int_a^b f(x) \mathrm{d} x=\int_a^c f(x) \mathrm{d} x+\int_c^b f(x) \mathrm{d} x .
\)

Example 4:

\(
\text { If } \int_0^8 f(x) d x=10 \text { and } \int_0^5 f(x) d x=5 \text {, find the value of } \int_5^8 f(x) d x
\)

We know that

\(
\begin{aligned}
\int_a^b f(x) d x &=\int_a^c f(x) d x+\int_c^b f(x) d x . \\
\int_0^8 f(x) d x &=\int_0^5 f(x) d x+\int_5^8 f(x) d x \\
10 &=5+\int_5^8 f(x) d x \\
5 &=\int_5^8 f(x) d x
\end{aligned}
\)

 

8. If \(f(x) \geq 0\), then \(\int_a^b f(x) \mathrm{d} x \geq 0\)

Example 5:

In this example \(f(x)\) = 4 \(x\), As \(f(x) \geq 0 \) its integral is also \(\int_1^2 4 x d x \geq 0 \)

\(
\begin{aligned}
\int_1^2 4 x d x &=2 \left[x^2\right]_1^2 \\
&= 2(2^2-1^2) \\
&=6
\end{aligned}
\)

9. If \(f(x) \geq g(x)\), then \(\int_a^b f(x) \mathrm{d} x \geq \int_a^b g(x) \mathrm{d} x .\)

Example 6: 

\(
\text { Compare } f(x)=\sqrt{1+x^2} \text { and } g(x)=\sqrt{1+x} \text { over the interval }[0,1]
\)
Graphing these functions is necessary to understand how they compare over the interval \([0,1]\). Initially, when graphed on a graphing calculator, \(f(x)\) appears to be above \(g(x)\) everywhere. However, on the interval \([0,1]\), the graphs appear to be on top of each other. We need to zoom in to see that, on the interval \([0,1], g(x)\) is above \(f(x)\). The two functions intersect at \(x=0\) and \(x=1\) (as shown in figure a & b).

We can see from the graph that over the interval \([0,1], g(x) \geq f(x)\). Comparing the integrals over the specified interval \([0,1]\), we also see that \(\int_0^1 g(x) d x \geq \int_0^1 f(x) d x\) (Figure c & d). The thin, red-shaded area shows just how much difference there is between these two integrals over the interval \([0,1]\).

10. We also have the bounded property; if \(m \leq f(x) \leq M\), then
\(
m(b-a) \leq \int_a^b f(x) \mathrm{d} x \leq M(b-a)
\)

Example 7:

Suppose that on \([-2,5]\), the values of \(f\) lie in the interval \([m, M]\). Between which bounds does \(\int_{-2}^5 f(x) \mathrm{d} x\) lie?

In this example, we want to find the bounds of an integral using the property where the values of \(f\) lie in a particular interval.
In particular, the following property states that if \(m \leq f(x) \leq M\), then
\(
m(b-a) \leq \int_a^b f(x) \mathrm{d} x \leq M(b-a)
\)
On applying this property with \(a=-2\) and \(b=5\), we have \(b-a=7\). Thus,
\(
7 m \leq \int_{-2}^5 f(x) \mathrm{d} x \leq 7 M
\)

11. The modulus property is given by
\(
\left|\int_a^b f(x) \mathrm{d} x\right| \leq \int_a^b|f(x)| \mathrm{d} x .
\)

Proof: 

From Negative of Absolute Value, we have for all \(a \in[a \ldots b]\) :
\(
-|f(t)| \leq f(t) \leq|f(t)|
\)
Thus from Relative Sizes of Definite Integrals:
\(
-\int_a^b|f(t)| \mathrm{d} t \leq \int_a^b f(t) \mathrm{d} t \leq \int_a^b|f(t)| \mathrm{d} t
\)
Hence the result.

12. Finally, we have a property for even and odd functions when integrating over the interval \([-a, a]\). For even functions \(f(-x)=f(x)\), we have
\(
\int_{-a}^a f(x) \mathrm{d} x=2 \int_0^a f(x) \mathrm{d} x .
\)

Example 8:

Evaluate \(\int_{-\pi / 2}^{\pi / 2} \cos ^3 x d x\)

Solution:
\(
\begin{aligned}
&\text { Let } f(x)=\cos ^3 x \\
&f(x)=(\cos x)^3 \\
&f(-x)=(\cos (-x))^3 \\
&f(-x)=(\cos x)^3 \\
&f(-x)=\cos ^3 x \\
&f(-x)=f(x)
\end{aligned}
\)
The function \(f(x)\) is even.
\(
\begin{aligned}
\int_{-\pi / 2}^{\pi / 2} \cos ^3 x d x &=2 \int_0^{\pi / 2} \cos ^3 x d x \\
&=2 \int_0^{\pi / 2} \frac{(\cos 3 x+3 \cos x)}{4} \\
&=(1 / 2)\left[\frac{\sin 3 x}{3}+3 \sin x\right]_0^{\pi / 2} \\
&=(1 / 2)[(-1 / 3)+3)] \\
&=(1 / 2)(8 / 3) \\
&=4 / 3
\end{aligned}
\)

For odd functions \(f(-x)=-f(x)\), we have
\(
\int_{-a}^a f(x) \mathrm{d} x=0
\)

Example 9:

Evaluate \(\int_{-\pi / 4}^{\pi / 4} x^3 \cos ^3 x d x\)

Solution:
\(
\begin{aligned}
&\int^{\pi / 4} x^3 \cos ^3 x d x \\
&-\pi / 4 \\
&f(x)=x^3 \cos ^3 x \\
&f(x)=(x \cos x)^3 \\
&f(-x)=(-x \cos (-x))^3 \\
&f(-x)=-x \cos ^3 x
\end{aligned}
\)
The given function is odd.
\(
\int_{-\pi / 4}^{\pi / 4} x^3 \cos ^3 x d x=0
\)

13. \(\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\)

Proof: Let \(u=a+b-x\)
\(
\Rightarrow d u=-d x
\)
\(
\int_a^b f(x) d x=\int_a^b f(u) d u
\)
\(
\begin{aligned}
&\Rightarrow \int_b^a f(a+b-x)(-d x) \\
&\Rightarrow-\int_b^a f(a+b-x) d x \\
&\Rightarrow \int_a^b f(a+b-x) d x
\end{aligned}
\)
Hence Proved.

14. \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)

Proof:

\(I=\int_0^a f(x) d x\)
now, put \(a-t=x; -d t=d x\)
also when,
\(
\begin{aligned}
&x=0; t=a \\
&x=a; t=0
\end{aligned}
\)
\(
\begin{array}{rlr}
I=\int_a^0 f(a-t)(-d t)=\int_0^a f(a-t) d t \\
=\int_0^a f(a-x) d x 
\end{array}
\)

Hence proved. The dummy variable does not affect the definite integral. So replacing \(t\) with \(x\) does not make any difference

15. \(\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-x) d x\)

Proof:

\(I=\int_0^{2a} f(x) d x=\int_0^a f(x) d x + \int_a^{2 a} f(x) d x\)
now, put \(2a-x=t; -d t=d x\)
also when,
\(
\begin{aligned}
&x=a; t=a \\
&x=2a; t=0
\end{aligned}
\)

\(
\begin{array}{rlr}
I=\int_0^{2a} f(x) d x=\int_0^a f(x) d x + \int_a^0 f(2a-t) (-d t) \\
=\int_0^a f(x) d x +\int_0^a f(2a-t) d t \\
=\int_0^a f(x) d x +\int_0^a f(2a-x) d x
\end{array}
\)

Hence proved.

16. \(
\int_0^{2 a} f(x) d x=\left\{\begin{array}{l}
2 \int_0^a f(x) d x, \text { if } f(2 a-x)=f(x) \\
0, \text { if } f(2 a-x)=-f(x)
\end{array}\right.
\)

Proof:

Let’s use the following substitution

put \(2a-x=t; -d t=d x\)
also when,
\(
\begin{aligned}
&x=a; t=a \\
&x=2a; t=0
\end{aligned}
\)

\(
\begin{array}{rlr}
I=\int_0^{2a} f(x) d x =\int_0^{a} f(x) d x + \int_a^{2a} f(x) d x [x = (2a-t)] \\
=\int_0^a f(x) d x + \int_a^0 f(2a-t) (-d t) \\
=\int_0^a f(x) d x +\int_0^a f(2a-t) d t \\
=\int_0^a f(x) d x +\int_0^a f(2a-x) d x \\
=2\int_0^{2a} f(x) d x (f(2 a-x)=f(x))
\end{array}
\)

 

\(
\begin{array}{rlr}
I=\int_0^{2a} f(x) d x=\int_0^a f(x) d x + \int_a^0 f(2a-t) (-d t) \\
=\int_0^a f(x) d x +\int_0^a f(2a-t) d t \\
=\int_0^a f(x) d x +\int_0^a f(2a-x) d x \\
=\int_0^a f(x) d x -\int_0^a f(x) d x = 0  [f(2a-x) = -f(x)]
\end{array}
\)