We will be exploring some of the important properties of definite integrals and their proofs in this article to get a better understanding. Integration is the estimation of an integral. It is just the opposite process of differentiation.Â Integrals in maths are used to find many useful quantities such as areas, volumes, displacement, etc.Â There are two types of Integrals namely, definite integral and indefinite integral. Here, we will learn about definite integrals and its properties, which will help to solve integration problems based on them.
Table of Contents:
Definite Integral Definition
If an integral has upper and lower limits, it is called a Definite Integral. There are many definite integral formulas and properties.Â Definite Integral is the difference between the values of the integral at the specified upper and lower limit of the independent variable. It is represented as;
âˆ«_{a}^{b} f(x) dx
Definite Integral Properties
Following is the list of definite integrals in the tabular form which is easy to read and understand.
Properties 
Description 
Property 1 
_{p}âˆ«^{q}Â f(a) da = _{p}âˆ«^{q}Â f(t) dt 
Property 2  _{p}âˆ«^{q} f(a) d(a) = – _{q}âˆ«^{p}Â f(a) d(a), Also _{p}âˆ«^{p}Â f(a) d(a) = 0 
Property 3  _{p}âˆ«^{qÂ }f(a) d(a) = _{p}âˆ«^{r}Â f(a) d(a) + _{r}âˆ«^{q}Â f(a) d(a) 
Property 4  _{p}âˆ«^{q} f(a) d(a) = _{p}âˆ«^{q}Â f( p + q – a) d(a) 
Property 5  _{o}âˆ«^{p} f(a) d(a) = _{o}âˆ«^{p}Â f(p – a) d(a) 
Property 6  âˆ«_{0}^{2p} f(a)da = âˆ«_{0}^{pÂ }f(a)da +âˆ«_{0}^{pÂ }f(2pa)da…if f(2pa) = f(a) 
Property 7  2 parts

Property 8  2 parts

Properties of Definite Integrals Proofs
Property 1: _{p}âˆ«^{q}Â f(a) da = _{p}âˆ«^{q}Â f(t) dt
This is the simplest property as only a is to be substituted by t, and the desired result is obtained.
Property 2: _{p}âˆ«^{q} f(a) d(a) = – _{q}âˆ«^{p}Â f(a) d(a), Also _{p}âˆ«^{p}Â f(a) d(a) = 0
Suppose I = _{p}âˆ«^{q} f(a) d(a)
If fâ€™ is the antiderivative of f, then use the second fundamental theorem of calculus, to get I = fâ€™(q)fâ€™(p) = – [fâ€™(p) – fâ€™(q)] = – _{q}âˆ«^{p}(a)da. Also, if p = q, then I= fâ€™(q)fâ€™(p) = fâ€™(p) fâ€™(p) = 0. Hence, _{a}âˆ«^{a}f(a)da = 0.
Property 3: _{p}âˆ«^{qÂ }f(a) d(a) = _{p}âˆ«^{r}Â f(a) d(a) + _{r}âˆ«^{q}Â f(a) d(a)
If fâ€™ is the antiderivative of f, then use the second fundamental theorem of calculus, to get;
_{p}âˆ«^{qÂ }f(a)da = fâ€™(q)fâ€™(p)… (1)
_{p}âˆ«^{r}f(a)da = fâ€™(r) – fâ€™(p)… (2)
_{r}âˆ«^{q}f(a)da = fâ€™(q) – fâ€™(r) â€¦ (3)
Letâ€™s add equations (2) and (3), to get
_{p}âˆ«^{rÂ }f(a)daf(a)da + _{r}âˆ«^{qÂ }f(a)daf(a)da = fâ€™(r) – fâ€™(p) + fâ€™(q)
= fâ€™(q) – fâ€™(p) = _{p}âˆ«^{qÂ }f(a)da
Property 4: _{p}âˆ«^{q} f(a) d(a) = _{p}âˆ«^{q}Â f( p + q – a) d(a)
Let, t = (p+qa), or a = (p+q – t), so that dt = – da … (4)
Also, note that when a = p, t = q and when a = q, t = p. So, _{p}âˆ«^{q} wil be replaced by _{q}âˆ«^{p} when we replace a by t. Therefore,
_{p}âˆ«^{qÂ }f(a)da = –_{q}âˆ«^{pÂ }f(p+qt)dt â€¦ from equation (4)
From property 2, we know that _{p}âˆ«^{qÂ }f(a)da = – _{q}âˆ«^{pÂ }f(a)da. Use this property, to get
_{p}âˆ«^{qÂ }f(a)da =_{p}âˆ«^{qÂ }f(p+qt)da
Now use property 1 to get
_{p}âˆ«^{qÂ }f(a)da = _{p}âˆ«^{qÂ }f(p + q – a )da
Property 5:Â \(\int_{0}^{p}\)f(a)da = \(\int_{0}^{p}\)f(pa)da
Let, t = (pa) or a = (p – t), so that dt = – da â€¦(5)
Also, observe that when a = 0, t =p and when a = p, t = 0. So, \(\int_{0}^{p}\) will be replaced by \(\int_{o}^{p}\) when we replace a by t. Therefore,
\(\int_{0}^{p}\)f(a)da = – \(\int_{p}^{0}\)f(p – t)da â€¦ from equation (5)From Property 2, we know that \(\int_{p}^{q}\)f(a)da = \(\int_{q}^{p}\)f(a)da. Using this property , we get
\(\int_{0}^{p}\)f(a)da = \(\int_{0}^{p }\)f(pt)dtNext, using Property 1, we get
\(\int_{0}^{a}\)f(a)da = \(\int_{0}^{p}\)f(p – a)da
Property 6:Â \(\int_{0}^{2p}\)f(a)da = \(\int_{0}^{p}\)f(a)da + \(\int_{0}^{p}\)f(2p – a))da
From property 3, we know that
\(\int_{p}^{q}\)f(a)da = \(\int_{p}^{r}\)f(a)da + \(\int_{r}^{q}\)f(a)daTherefore, \(\int_{0}^{2p}\)f(a)da = \(\int_{0}^{p}\)f(a)da + \(\int_{p}^{2p}\)f(a)da = I_{1} + I_{2} â€¦ (6)
Where, I_{1} = \(\int_{0}^{p}\)f(a)da and I_{2} =\(\int_{p}^{2p}\)f(a)da
Let, t = (2p – a) or a = (2p – t), so that dt = da â€¦(7)
Also, note that when a = p, t = p, and when a =2p, t= 0. Hence, \(\int_{a}^{0}\) when we replace a by t. Therefore,
I_{2 } = \(\int_{p}^{2p}\)f(a)da = – \(\int_{p}^{0}\)f(2p0)daâ€¦ from equation (7)
From Property 2, we know that \(\int_{p}^{q}\)f(a)da = \(\int_{q}^{p}\)f(a)da. Using this property, we get I_{2} = \(\int_{0}^{p}\)f(2pt)dt
Next, using Property 1, we get
I_{2 = }\(\int_{0}^{a}\)f(a)da + \(\int_{0}^{a}\)f(2pa)da
Replacing the value of I_{2 } in equation (6), we get
\(\int_{0}^{2p}\)f(a)da = \(\int_{0}^{p}\)f(a)da + \(\int_{o}^{p}\)f(2p – a)da
Property 7:Â \(\int_{0}^{2a}\)f(a)da = 2 \(\int_{0}^{a}\)f(a)da â€¦ if f(2p – a) = f(a) and
\(\int_{0}^{2a}\)f(a)da = 0 â€¦ if f(2p a) = f(a)
we know that
\(\int_{0}^{2p}\)f(a)da =\(\int_{0}^{p}\)f(a)da + \(\int_{0}^{p}\)f(2p – a)da â€¦ (8)Now, if f(2p – a) = f(a), then equation (8) becomes
\(\int_{0}^{2p}\)f(a)da = \(\int_{0}^{p}\)f(a)da + \(\int_{0}^{p}\)f(a)da=2\(\int_{0}^{p}\)f(a)da
And, if f(2p – a) = – f(a), then equation (8) becomes
\(\int_{0}^{2p}\)f(a)da = \(\int_{0}^{p}\)f(a)da \(\int_{0}^{p}\)f(a)da = 0
Property 8:Â \(\int_{p}^{p}\)f(a)da = 2\(\int_{0}^{p}\)f(a)da â€¦ if f(a) =f(a) or it is an even function and \(\int_{a}^{a}\)f(a)da = 0, â€¦ if f(a) = f(a) or it is an odd function.
Using Property 3, we have
\(\int_{p}^{p}\)f(a)da = \(\int_{a}^{0}\)f(a)da + \(\int_{0}^{p}\)f(a)da = I_{1 }+ I,_{2} â€¦(9)Where, I_{1 =}\(\int_{a}^{0}\)f(a)da, I_{2} =\(\int_{0}^{p}\)f(a)da
Consider I_{1 }
Let, t = a or a = t, so that dt = dx â€¦ (10)
Also, observe that when a = p, t = p, when a = 0, t =0. Hence, \(\int_{a}^{0}\) will be replaced by \(\int_{a}^{0}\) when we replace a by t. Therefore,
I_{1} = \(\int_{a}^{0}\)f(a)da = – \(\int_{a}^{0}\)f(a)da â€¦ from equation (10)
From Property 2, we know that\(\int_{p}^{q}\)f(a)da = – \(\int_{q}^{p}\)f(a)da, use this property to get,
I_{1 =}\(\int_{p}^{0}\)f(a)da = \(\int_{0}^{p}\)f(a)da
Next, using Property 1, we get
I_{1} = \(\int_{p}^{0}\)f(a)da = \(\int_{0}^{p}\)f(a)da
Replacing the value of I_{2} in equation (9), we get
\(\int_{p}^{p}\)f(a)da = I_{1} + I_{2} = \(\int_{0}^{p}\)f(a)da + \(\int_{0}^{p}\)f(a)da =2 \(\int_{0}^{p}\)f(a)da â€¦ (11)Now, if â€˜fâ€™ is an even function, then f(â€“ a) = f(a). Therefore, equation (11) becomes
\(\int_{p}^{p}\)f(a)da = \(\int_{0}^{p}\)f(a)da +\(\int_{0}^{p}\)f(a)da =2\(\int_{0}^{p}\)f(a)daAnd, if â€˜fâ€™ is an odd function, then f(â€“a) = â€“ f(a). Therefore, equation (11) becomes
\(\int_{p}^{p}\)f(a)da = – \(\int_{0}^{a}\)f(a)da + \(\int_{0}^{p}\)f(a)da = 0Now, let us evaluate Definite Integral through a problem sum.
Example
Question: Evaluate \(\int_{1}^{2}\)f(a^{3 }– a)da
Solution: Observe that, (a^{3} â€“ a) â‰¥ 0 on [â€“ 1, 0], (a^{3} â€“ a) â‰¤ 0 on [0, 1] and (a^{3} â€“ a) â‰¥ 0 on [1, 2]
Hence, using Property 3, we can write
\(\int_{1}^{2}\)f(a^{3 }– a)da =\(\int_{1}^{0}\)f(a^{3 }– a)da + \(\int_{0}^{1}\)f(a^{3 }– a)da + \(\int_{1}^{2}\)f(a^{3 }– a)da = \(\int_{1}^{0}\)f(a^{3 }– a)da +\(\int_{0}^{1}\)f(a – a^{3 })da + \(\int_{1}^{2}\)f(a^{3 }– a)da \(\int_{1}^{0}\)f(a^{3 }– a)da +\(\int_{0}^{1}\)f(a – a^{3 })da +\(\int_{1}^{2}\)f(a^{3 }– a)daSolving the integrals, we get
\(\int_{1}^{2}f(a3 – a)da = \frac{x4}{4} – (\frac{x2}{2}) ]10 + [(\frac{x2}{2} – (\frac{x4}{4}))01 + [ \frac{x4}{4} (\frac{x2}{2}) ]12\)= – [\(\frac{1}{4}\) – \(\frac{1}{2}\)] + [\(\frac{}{}\) – \(\frac{1}{4}\)] + [ 4 – 2] [\(\frac{1}{4}\) \(\frac{1}{2}\) = \(\frac{11}{4}\)
Subscribe to BYJUâ€™S to watch an explanatory video on Definite Integral and many more Mathematical topics.