Properties of Definite Integrals: Definition and Proof

Definite integral is an integral having two predefined limits, namely, upper limit and lower limit. Integral is the opposite or inverse of differentiation. It is also termed as anti-derivative. The process of estimating an integral is called integration. The concepts of integral are helpful to evaluate the area, volume, displacement, etc. Integrals Calculus is of two types – definite integrals and indefinite integrals. As the name suggests, definite integrals are those that have upper and lower limits. Whereas, indefinite integrals have no limit and include an arbitrary constant.

Read More: Methods of Integration

Key Terms: Definite Integral, Differentiation, Integration, Calculus, Indefinite Integral, Antiderivative 

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Definition of Definite Integral

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Definite integral has both upper and lower limits. Definite integrals is denoted by abf(x)dx, here, ‘b’ and ‘a’ are the upper and lower limit respectively. There are two ways to estimate the definite integral:

1. Definite integral as the limit of the sum
2. abf(x)dx=Fb-Fa, Where F is an anti-derivative of f(x).

Definite integral as the limit of the sum: The definite integral abf(x)dx denotes the area covered by the curve y which is equal to f(x). Here, the ordinates x=a, x=b, and the x-axis and denoted by:

OR

Definite Integrals Video Lecture:

Read More: Applications of Integrals

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Properties of Definite Integral

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The properties of definite integrals can be used to integrate the provided function and apply the lower and upper bounds to get the integral's value. The definite integral formulas can be used to calculate the integral of a function multiplied by a constant, the sum of the functions, and the integral of even and odd functions.

Property 1:

P₀: abfxdx= abf tdt

Property 2:

P₁: abfxdx= -abf xdx,  where, abf xdx=0

Property 3:

P₂: abfxdx= acfxdx+ cbfxdx

Property 4:

P₃: abfxdx= abf a+b-xdx

Property 5:

P₄: 0afxdx= 0af a-xdx

Property 6:

P₅: 02afxdx= 0af xdx+ 0af2a-xdx

Property 7:

P₆: 02afxdx= 20af xdx, if f2a-x=fx

AND

P₆: 02afxdx= 0, if f2a-x=-fx

Property 8:

P₇: 

• -aafxdx= 20af xdx,  if f is an even function i.e., f (-x) = f (x)
• -aafxdx= 0 , if f is an odd function i.e. , f (-x) = -f (x)

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Properties of Definite Integral Proofs

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Property 1:

P₀: abfxdx= abf tdt

Proof: abfxdx …….(1)

Assume, x = y

dx = dy

Putting the value in equation (1)

abfydy

Property 2:

P₁: abfxdx= -abf xdx,  where, abf xdx=0

Proof:

Let I =ab f(x) dx. Here, ‘F’ is anti-derivative of ‘f’, then by using the second fundamental theorem of calculus, we get

 I = F (b) – F (a) = – [F (a) – F (b)] = – ba f (x) dx.

Also, if a = b, then I = F (b) – F (a) = F (a) – F (a) = 0.

Hence, aaf(x) dx = 0.

Property 3:

P₂: abfxdx= acfxdx+ cbfxdx

Proof:

Here, ‘F’ is the anti-derivative of ‘f’, using the second fundamental theorem of calculus, we get

• abf(x) dx = F(b) – F(a) … (1)
• acf(x)  dx = F(c) – F(a) … (2)
• cbf(x)  dx = F(b) – F(c) … (3)

Addition of equations (2) and (3), results

acf(x) dx + bcf(x)  dx = F (c) – F (a) + F (b) – F (c)

= F (b) – F (a)

= abf(x)  dx

Read More:  Differential Equations

Property 4:

P₃: abfxdx= abf a+b-xdx

Proof:

Assume a + b - x = y………… (1)

-dy = dx

From equation (1), when x=a,

y = a + b – a

y = b

When, x=b,

y = a + b – b

y = a

Putting these values in the R.H.S. of the integral, we get

-bafydy

Using property 1 and property 2, we can say that

abfx dx= abfa+b-x dx

Property 5:

P₄: 0afxdx= 0af a-xdx

Proof: Assume, t = (a - x) or x = (a - t), we get dt = - dx……. (1)

Also, note that when x = 0, t = a, and when x = a, t = 0.

Hence, when x replaced by t, 0a will be replaced by a0

Using equation (1) we get,

 0a f xdx=- a0 fa-t dt

Using property 2, we get

0a fx dx= 0afa-t dt

Using property 1, we get

0a fx dx= 0a fa-x dx

Read More: Integration By Parts

Property 6:

P₅: 02afxdx= 0af xdx+ 0af2a-xdx

Proof:

abfxdx= acfxdx+ cbfxdx …….Property 3

Thus, 02af(x) dx = 0af(x)  dx + a2af(x) dx

= I₁ + I₂ ……. (1)

Here, I₁ = 0afxdx and I₂ = 02afxdx

Assume, t = (2a - x) or x = (2a - t),

We get, dt = - dx ……. (2)

Also, note that, when x = a, t = a, and when x = 2a, t = 0.

Thus, when we replace x by t, a2a will become a0

Therefore, I₂ =a2a fx dx= -a0f2a-t dt ……..from equation (2)

Using property 2, we get

I₂ = 0a f2a-t dt

Again, using property 1, we get

I₂ = 0a f2a-x dx

Putting value of I2, in equation (1), we get

02a fx dx = 0a fx dx+ 0af2a-x dx

Read More: Determinant Formula

Property 7:

P₆: 02afxdx= 20af xdx, if f2a-x=fx

AND

P₆: 02afxdx= 0, if f2a-x=-fx

Proof:

Using property 5, we get

02af(x) dx = 0af(x) dx + 0af(2a-x) dx …….(1)

If, f (2a-x) = f(x), then equation (1) becomes

02af(x) dx = 0afxdx + 0afxdx

= 2 0afx dx

If, f (2a-x) = - f(x), then equation (1) becomes

02afxdx = 0afxdx - 0afxdx

Thus, 0afxdx  = 0

Property 8:

P₇: 

1. -aafxdx= 20af xdx,  if f is an even function i.e., f (-x) = f (x)
2. -aafxdx= 0 , if f is an odd function i.e. , f (-x) = -f (x)

Proof:

Using property 3, we get

-aafxdx = -a0fxdx +0afxdx = I₁ + I₂ ……. (1)

Where, I₁ = -a0fxdx I2 = 0afxdx

Consider I₁, let t = -x, or say x = -t, thus, dt = -dx ……. (2)

Also, note that when x = -a, t = a, and when x = 0, t = 0.

When we replace x by t, -a0  will be replaced bya0 .

Thus, I₁ = -a0 fx dx= -a0 f-t dt .......using equation (2)

Using property 2, we have

I₁ = -a0 fxdx= 0af-t dt

Again, using property 1, we have

I₁ =-a0fxdx = 0afxdx

Putting the value of I₂ in equation (1), we get

-aafxdx = I₁ + I₂ = 0afxdx + 0afxdx ……. (3)

If ‘f’ is an even function, f (-x) = f(x). Thus, equation (3) becomes

-aafxdx =0afxdx +0afxdx = 2 0afxdx

And, if ‘f’ is an odd function, then f (– x) = – f(x). Therefore, equation (3) becomes

-aafxdx = – 0afxdx + 0afxdx = 0

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Things to Remember

• Definite integral formula is integral with upper and lower limits.
• There are two types of Integrals namely, definite integral and indefinite integral.
• Definite integrals can be used to determine the mass of an object if its density function is known.
• Difference between the values of the integral of a given function f(x) for an upper-value b and a lower value a of the independent variable x.
• Indefinite integral has an arbitrary constant. 

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Previous Years’ Questions 

1. Intersect the chord OP and the x-axis at points Q and R, respectively… [JEE Main – 2020]
2. The area of the region A = [(x,y):0… [JEE Main – 2019]
3. The area (in s units) of the region{(x,y):x… [JEE Main – 2017]
4. What will be the area enclosed between this curve and the coordinate axes… [JKCET – 2017]
5. The area (in s units) of the quadrilateral formed by the tangents at the endpoints… [JEE Main – 2015]
6. Let the straight line x=b divide the area enclosed by… [AMUEEE – 2014]
7. The area (in s units) bounded by the curves y=\(\sqrt{}\)x… [COMEDK UGET – 2013]
8. If S be the area of the region enclosed by… [JEE Advanced – 2012]
9. The area of the region between the curves y… [JEE Advanced – 2008]
10. Then sin2θ equals… [COMEDK UGET – 2005]

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