Definite Integrals of Piecewise Functions

Definite Integrals of Piecewise Functions

Komal MiglaniUpdated on 02 Jul 2025, 08:00 PM IST

Integration is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These integration concepts have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

Definite Integrals of Piecewise Functions
Definite Integrals of Piecewise Functions

Piecewise Definite integration

Definite integration calculates the area under a curve between two specific points on the x-axis.

Let f be a function of x defined on the closed interval [a, b]. F be another function such that $\frac{d}{d x}(F(x))=f(x)$ for all x in the domain of f, then $\int_a^b f(x) d x=[F(x)+c]_a^b=F(b)-F(a)$is called the definite integral of the function f(x) over the interval [a, b], where a is called the lower limit of the integral and b is called the upper limit of the integral.

Definite integrals have properties that relate to the limits of integration.

Property 1

If the upper and lower limits of integration are the same, the integral is just a line and contains no area, hence the value is 0. $\int_a^a f(x) d x=0$

Property 2

The value of the definite integral of a function over any particular interval depends on the function and the interval but not on the variable of the integration. $\int_a^b f(x) d x=\int_a^b f(t) d t=\int_a^b f(y) d y$

Property 3

If the limits of definite integral are interchanged, then its value changes by a minus sign only.

Property 4 (King's Property)

This is one of the most important properties of definite integration. $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

Property 5 (Piecewise Definite integration)

$\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{\mathrm{a}}^{\mathrm{c}} \mathrm{f}(\mathrm{x}) \mathrm{dx}+\int_{\mathrm{c}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ where $c \in \mathbb{R}$

This property is useful when the function is in the form of piecewise or discontinuous or non-differentiable at x = c in (a, b).
$
\begin{aligned}
& \text { Let } \quad \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{F}(\mathrm{x}))=\mathrm{f}(\mathrm{x}) \\
& \therefore \quad \int_a^c f(x) d x+\int_c^b f(x) d x \\
& =\left.\mathrm{F}(\mathrm{x})\right|_{\mathrm{a}} ^{\mathrm{c}}+\left.\mathrm{F}(\mathrm{x})\right|_{\mathrm{c}} ^{\mathrm{b}} \\
& =\mathrm{F}(\mathrm{c})-\mathrm{F}(\mathrm{a})+\mathrm{F}(\mathrm{b})-\mathrm{F}(\mathrm{c}) \\
& =\mathrm{F}(\mathrm{b})-\mathrm{F}(\mathrm{a}) \\
& =\int_a^b f(x) d x
\end{aligned}
$

The above property can be also generalized into the following form

$\int_a^b f(x) d x=\int_a^{c_1} f(x) d x+\int_{c_1}^{c_2} f(x) d x+\ldots+\int_{c_n}^b f(x) d x$

Property 6

$\int_0^a f(x) d x=\int_0^{a / 2} f(x) d x+\int_0^{a / 2} f(a-x) d x$

Proof:

From the previous property,

$
\int_0^a f(x) d x=\int_0^{a / 2} f(x) d x+\int_{a / 2}^a f(x) d x
$

Put $x=a-t \Rightarrow d x=-d t$ in the second integral, when $x=a / 2$, then $t=a / 2$ and when $x=a$, then $t=0$

$
\therefore \quad \begin{aligned}
\int_0^a f(x) d x & =\int_0^{a / 2} f(x) d x+\int_{a / 2}^0 f(a-t)(-d t) \\
& =\int_0^{\mathrm{a} / 2} \mathrm{f}(\mathrm{x}) \mathrm{dx}+\int_0^{\mathrm{a} / 2} \mathrm{f}(\mathrm{a}-\mathrm{t}) \mathrm{dt} \\
\int_0^a f(x) d x & =\int_0^{a / 2} f(x) d x+\int_0^{a / 2} f(a-x) d x
\end{aligned}
$

Recommended Video Based on Piecewise Definite Integration


Solved Examples Based on Piecewise Definite Integration:

Example 1: The integral $\int_0^\pi \sqrt{1+4 \sin ^2 \frac{x}{2}-4 \sin \frac{x}{2}} d x$ equals:

1) $4 \sqrt{3}-4$
2) $4 \sqrt{3}-4-\frac{\pi}{3}$
3) $\pi-4$
4) $\frac{2 \pi}{3}-4-4 \sqrt{3}$

Solution

As learnt in concept

Fundamental Properties of Definite integration -

If the function is continuous in (a, b ) then integration of a function a to b will be same as the sum of integrals of the same function from a to c and c to b.

$\int_b^a f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x$

- wherein

$\begin{aligned} & =\int_0^\pi \sqrt{\left(1-2 \sin \frac{x}{2}\right)^2}=\int_0^\pi\left|1-2 \sin \frac{x}{2}\right| d x \\ & \int_0^{\frac{\pi}{3}} \sqrt{\left(1-2 \sin \frac{x}{2}\right)}=\int_{\frac{\pi}{4}}^\pi\left(1-2 \sin \frac{x}{2}\right) d x \\ & \left(x+4 \cos \frac{x}{2}\right)_0^{\frac{\pi}{3}}-\left(x+4 \cos \frac{x}{2}\right)^\pi \\ & =4 \sqrt{3}-4-\frac{\pi}{3}\end{aligned}$

Example 2: The value of $\int_1^a[x] f^{\prime}(x) d x, a>1$, where $[x]$ denotes the greatest integer not exceeding $x$ is
1) $a f(a)-\{f(1)+f(2)+\ldots+f([a])\}$
2) $[a] f(a)-\{f(1)+f(2)+\ldots+f([a])\}$
3) $[a] f([a])-\{f(1)+f(2)+\ldots+f(a)\}$
4) $a f([a])-\{f(1)+f(2)+\ldots+f(a)\}$

Solution

As learnt in concept

Fundamental Properties of Definite integration -

If the function is continuous in (a, b ) then integration of a function a to b will be same as the sum of integrals of the same function from a to c and c to b.

$\int_b^a f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x$

- wherein


$[x]$ has to be split into integral limits.

$\int_1^a[x] f^{\prime}(x) d x$

=$\int_1^2 f^{\prime}(x) d x+\int_2^3 2 f^{\prime}(x) d x+\ldots-\ldots+\int_{[a]}^a[a] f^{\prime}(x) d x$

$\begin{aligned} & f(2)-f(1)+2 f(3)-2 f(2)+--------------- \\ = & ---------+[a] f(a)-[a] f([a])\end{aligned}$

Terms start cancelling out,

We get,

$\begin{aligned} & -f(1)-f(2)-f(3)------------------- \\ - & ---f[a]+[a] f(a)\end{aligned}$

=$=[a] f(a)-(f(1)+f(2)+\cdots-\cdots-\cdots-\cdots-\cdots([a])$

Example 3: $\int_0^{\sqrt{2}}\left[x^2\right] d x$ is

1) $2-\sqrt{2}$
2) $2+\sqrt{2}$
3) $\sqrt{2}-$
4) $\sqrt{2}-2$

Solution

Fundamental Properties of Definite integration -

If the function is continuous in (a, b ) then integration of a function a to b will be same as the sum of integrals of the same function from a to c and c to b.

$\int_b^a f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x$

- wherein

$\begin{aligned} & \int_0^2\left[x^2\right] d x=\int_0^1\left[x^2\right] d x+\int_1^{\sqrt{2}}\left[x^2\right] d x \\ & \Rightarrow 0+\int_1^{\sqrt{2}} 1 d x=\sqrt{2}-1\end{aligned}$

Example 4: Choose the correct option?

1) $\int_{-a}^a f(x) d x=\int_{-a}^0 f(x) d x+\int_0^a f(x) d x$
2) $\int_a^c f(x) d x=\int_a^d f(x) d x+\int_d^c f(x) d x$; where $a<d<c$
3) $\int_a^c f(x) d x=\int_a^b f(x) d x+\int_b^c f(x) d x$; where $a<c<b$

4) All are true.

Solution

As we have learnt,

Fundamental Properties of Definite integration -

If the function is continuous in (a, b ) then integration of a function a to b will be same as the sum of integrals of the same function from a to c and c to b.

$\int_b^a f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x$

- wherein

It doesn't matter if b lies between a and c or not.


Example 5: The value of the integral $\int_{-2}^2 \frac{\sin ^2 x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}} d x$(where $[x]$ denotes the greatest integer less than or equal to (x) ) is :

1) $\sin 4$

2) 0

3) 4

4) $4-\sin 4$

Solution

Fundamental Properties of Definite integration -

If the function is continuous in (a, b ) then integration of a function a to b will be same as the sum of integrals of the same function from a to c and c to b.

$\int_b^a f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x$

- wherein


$\begin{aligned} & I=\int_2^{-2} \frac{\sin ^2 x}{\left|\frac{1}{7}\right|+\frac{1}{2}} d x \\ & =\int_2^0 \frac{\sin ^2 x}{\frac{1}{2}} d x+\int_0^{-2} \frac{\sin ^2 x}{-1+\frac{1}{2}} d x \\ & 2 \oint_2^0 \sin ^2 x d x 1-2 \int_0^{-2} \sin ^2 x d x \\ & P u t x=-p \Rightarrow d x=-d y \sin ^2(-p)=\sin ^2 p \\ & =2 \oint_0^2 \sin ^2 x d x+2 \int_2^0 \sin ^2 P d p\end{aligned}$

Frequently Asked Questions (FAQs)

Q: What is the relationship between the definite integral of a piecewise function and its indefinite integral?
A:
The relationship between the definite and indefinite integrals of a piecewise function is:
Q: How does the concept of accumulation functions apply to piecewise functions?
A:
Accumulation functions for piecewise functions work as follows:
Q: How do you find the area between two piecewise functions?
A:
To find the area between two piecewise functions f(x) and g(x):
Q: What is the significance of the Second Fundamental Theorem of Calculus for piecewise functions?
A:
The Second Fundamental Theorem of Calculus for piecewise functions states that:
Q: What is the relationship between the derivative and integral of a piecewise function?
A:
The relationship between the derivative and integral of a piecewise function is complex:
Q: How do you handle definite integrals of piecewise functions with absolute value components?
A:
To handle definite integrals of piecewise functions with absolute value components:
Q: What is the importance of checking for continuity at transition points when integrating piecewise functions?
A:
Checking for continuity at transition points is important because:
Q: How does the area interpretation of definite integrals apply to piecewise functions?
A:
The area interpretation of definite integrals applies to piecewise functions in a piecewise manner:
Q: How does integrating a piecewise function affect its continuity?
A:
Integrating a piecewise function generally increases its continuity:
Q: How do you find the average value of a piecewise function over a given interval?
A:
To find the average value of a piecewise function over an interval [a,b]: