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Applications of Definite Integrals

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  • Last Updated : 02 Feb, 2023
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Definite Integrals are used to find areas of the complex curve, volumes of irregular shapes, and other things. Definite Integrals are defined by, let us take p(x) to be the antiderivative of a continuous function f(x) defined on [a, b] then, the definite integral of f(x) over [a, b] is denoted by\int\limits_{a}^{b}f(x)dx  and is equal to [p(b) – p(a)].

\bold{\int\limits_{a}^{b}f(x)dx}   = p(b) – p(a)

The numbers a and b are called the limits of integration where a is called the lower limit and b is called the upper limit. The interval [a, b] is called the interval of the integration.

Note

  • Constant of Integration is not included in the evaluation of the definite integral.
  • \bold{\int\limits_{a}^{b}f(x)dx}   is read as “integral of f(x) from a to b”

How to Evaluate Definite Integrals?

To find the definite integral of f(x) over interval [a, b] i.e.,\int\limits_{a}^{b}f(x)dx   we have following steps:

Step 1: Find the indefinite integral ∫f(x) dx.

Step 2: Evaluate p(a) and p(b) where, p(x) is the antiderivative of f(x), p(a) is the value of antiderivative at x = a, and p(b) is the value of antiderivative at x = b.

Step 3: Calculate p(b) – p(a).

The value obtained in Step 3 is the desired value of the definite integral.

For more detail click here,

Properties of Definite Integral

Various Properties of Definite Integral have been discussed below:

Property 1)\bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(z)dz}

Proof:

Let p(x) be an antiderivative of f(x). Then,

\frac{d}{dx}   {p(x)} = f(x)

\frac{d}{dz}   {p(z)} = f(z)

\int\limits_{a}^b f(x)dx = \big[p(x)\big]_a^b   = p(b) – p(a) —————–(i)

\bold{\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx}

and\int\limits_{a}^b f(z)dz = \big[p(z)\big]_a^b   = p(b) – p(a) —————–(ii)

From (i) and (ii)

\bold{\int\limits_{a}^{b}f(x)dx= \int\limits_{a}^{b}f(z)dz}

Property 2)\bold{\int\limits_{a}^{b}f(x)dx=-\int\limits_{b}^{a}f(x)dx}

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

Proof:

Let p(x) be an antiderivative of f(x). Then,

\int\limits_{a}^b f(x)dx   = p(b) – p(a)

and-\int\limits_{b}^a f(x)dx   = -[p(a) – p(b)] = p(b) – p(a)

\bold{\int\limits_{a}^{b}f(x)dx=-\int\limits_{b}^{a}f(x)dx}

Property 3)\bold{\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx}   where, a<c<b

Proof:

Let p(x) be the antiderivative of f(x). Then,

\int\limits_{a}^b f(x)dx   = p(b) – p(a) —————(i)

\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx   = [p(c) – p(a)] + [p(b) – p(c)] = p(b) – p(a) —————(ii)

From (i) and (ii)

\bold{\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx}

Property 4)\bold{\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx}

Proof:

Let x = a – t. Then, dx = d(a – t) ⇒ dx = -dt

When x=0 ⇒ t = a and x = a ⇒ t = 0

\int\limits_{0}^{a}f(x)dx=-\int\limits_{a}^{0}f(a - t)dt

\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a - t)dt   [Using second property]

\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a - x)dx   [Using first property]

\bold{\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx}

Property 5)\bold{\int\limits_{-a}^{a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & \text{, if $n$ is even} \\ 0 & \text{, if $n$ is odd} \end{cases}}

Proof:

Using third property

\int\limits_{-a}^{a}f(x)dx=\int\limits_{-a}^{0}f(x)dx+\int\limits_{0}^{a}f(x)dx   ———-(i)

Let x = -t, dx = -dt

Limits: x = -a ⇒ t = a and x = 0 ⇒ t=0

\int\limits_{-a}^{0}f(x)dx=\int\limits_{a}^{0}f(-t)(-dt) =-\int\limits_{a}^{0}f(-t)dt=\int\limits_{0}^{a}f(-t)dt   [By second property]

\int\limits_{-a}^{0}f(x)dx=\int\limits_{0}^{a}f(-x)dx   [By first property] ————-(ii)

From (i) and (ii)

\int\limits_{-a}^{a}f(x)dx=\int\limits_{0}^{a}f(-x)dx+\int\limits_{0}^{a}f(x)dx

\int\limits_{-a}^{a}f(x)dx=\int\limits_{0}^{a}[f(-x)+f(x)]dx

\int\limits_{-a}^{a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(-x) = f(x) \\ 0 & , if f(-x) = -f(x) \end{cases}

\bold{\int\limits_{-a}^{a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & \text{, if $n$ is even} \\ 0 & \text{, if $n$ is odd} \end{cases}}

Property 6) If f(x) is a continuous function defined on [0, 2a]\bold{\int\limits_{0}^{2a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(2a - x) = f(x) \\ 0 & , if f(2a - x) = -f(x)\end{cases}}

Proof:

Using third property

\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^{a}f(x)dx+\int\limits_{a}^{2a}f(x)dx   ———–(i)

Consider

Let x = 2a – t, dx = -d(2a – t) ⇒ dx = -dt

Limits: x = a ⇒ t = a and x = 2a ⇒ t=0

\int\limits_{a}^{2a}f(x)dx=-\int\limits_{a}^{0}f(2a - t)dt

\int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a - t)dt   [Using second property]

\int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a - x)dx   [Using first property]

Substituting\int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a - x)dx   in (i)

\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^{a}f(x)dx+\int\limits_{0}^{a}f(2a - x)dx = \int\limits_{0}^{a}[f(x) + f(2a - x)]dx

\bold{\int\limits_{0}^{2a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(2a - x) = f(x) \\ 0 & , if f(2a - x) = -f(x)\end{cases}}

Property 7)\bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a + b - x)dx}

Proof:

Let t = a + b – x ⇒ dt = -dx

Limits: x = a, y=b and x = b, y =a

After putting value and limit of t in\int\limits_{a}^{b}f(a + b - x)dx

\int\limits_{a}^{b}f(a + b - x)dx =-\int\limits_{b}^{a}f(t)dt

\int\limits_{a}^{b}f(a + b - x)dx =\int\limits_{a}^{b}f(t)dt   [Using second property]

\int\limits_{a}^{b}f(a + b - x)dx =\int\limits_{a}^{b}f(x)dx   [Using first property]

\bold{\int\limits_{a}^{b}f(x)dx= \int\limits_{a}^{b}f(a + b - x)dx}

For more details on Properties of Definite Integral click here,

Definite Integral as Limit of Sum

The definite integral of f(x) over the interval [a, b], denoted by\int\limits_a^bf(x)dx   , is defined as the limit of a sum given by:

\int\limits_a^bf(x)dx = \displaystyle \lim_{n\to\infty}\displaystyle \sum_{r=1}^n hf(a + rh) \\ or \\ \displaystyle\lim_{n\to\infty}\displaystyle \sum_{r=0}^{n-1} hf(a + rh)

where nh = b – a

f(x) is said to be integrable over [a, b] if the above two limits exist and are equal.

Challenging Definite Integral

Problem1: Evaluate the definite integral:\int\limits^2_0 x^2[x]dx

Solution:

The integral contains the greatest integer.

For limit 0 to 1 greatest integer function [x] = 0

For limit 1 to 2 greatest integer function [x] = 1

So, we split the above integral into two parts using the following definite integral property.

\int\limits_a^b f(x)dx = \int\limits^c_af(x)dx + \int\limits^b_cf(x)dx

\int\limits_0^2x^2[x]dx = \int\limits_0^1x^2[x]dx+\int\limits_1^2x^2[x]dx

=\int\limits_0^1x^2.0dx+\int\limits_1^2x^2.1dx

\int\limits_0^2x^2[x]dx = 0+\int\limits_1^2x^2dx

\int\limits_0^2x^2[x]dx = \int\limits_1^2x^2dx

\int\limits_0^2x^2[x]dx = \big[\frac{x^3}{3}\big]_1^2

= [8/3] – [1/3]

\int\limits_0^2x^2[x]dx   = 7/3

Problem 2: Evaluate:\int\limits^3_1|x^2-2x| dx

Solution:

The integral contains a mod function.

For limit 1 to 2 let x=1, the function f(x) = x2 – 2x = 1-2 = -1 is negative.

For limit 2 to 3 let x=3, the function f(x) = x2 – 2x = 9-6 = 3 is positive.

So, we split the above limit of integral into two parts using the following definite integral property.

\int\limits_a^b f(x)dx = \int\limits^c_af(x)dx + \int\limits^b_cf(x)dx

\int\limits^3_1|x^2-2x| dx =\int\limits^2_1|x^2-2x| dx +\int\limits^3_2|x^2-2x| dx

\int\limits^3_1|x^2-2x| dx =\int\limits^2_1-(x^2-2x) dx +\int\limits^3_2(x^2-2x) dx

\int\limits^3_1|x^2-2x| dx =\int\limits^2_1(-x^2+2x) dx +\int\limits^3_2(x^2-2x) dx

\int\limits^3_1|x^2-2x| dx =\big[x^2-\frac{x^3}{3}\big]^2_1 +\big[\frac{x^3}{3}-x^2\big]^3_2

\int\limits^3_1|x^2-2x| dx =\big[4-\frac{8}{3}\big]-\big[1-\frac{1}{3}\big] +\big[9-9\big]-\big[\frac{8}{3}-4\big]

\int\limits^3_1|x^2-2x| dx   = 2/3 + 4/3

\int\limits^3_1|x^2-2x| dx   = 2

Problem 3: Evaluate:\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx

Solution:

\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx

Let √x = t

1/ (2√x) dx = dt

dx/√x = 2dt

Limits: If x=0, t=0 and x=π2/4, t=π/2

\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx = 2\int\limits_0^{\pi/2}costdt

\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx = 2\big[sint\big]_0^{\pi/2}

\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx   = 2(1 – 0)

\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx   = 2

Problem 4: Evaluate:\int\limits^{\pi /3}_{\pi/6}\frac{1}{1+\sqrt {cotx}}dx

Solution:

I =\int\limits^{\pi /3}_{\pi/6}\frac{1}{1+\sqrt {cotx}}dx

I =\int\limits^{\pi /3}_{\pi/6}\frac{1}{1+\frac{\sqrt {sinx}}{\sqrt{cosx}}}dx

I =\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx   ———-(1)

Using property\int\limits^b_af(x)dx = \int\limits^b_af(a+b-x)dx

I =\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{sin(\frac{\pi}{2}-x)}}{\sqrt{cos(\frac{\pi}{2}-x)}+\sqrt {sin(\frac{\pi}{2}-x)}}dx

I =\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt {cosx}}dx   ———-(2)

Adding (1) and (2)

2I =\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx+\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt {cosx}}dx

2I =\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{sinx}+\sqrt{cosx}}{\sqrt{sinx}+\sqrt {cosx}}dx

2I =\int\limits^{\pi /3}_{\pi/6}1.dx

2I =\big[x\big]^{\pi/3}_{\pi/6}

2I = (π/3) – (π/6)

2I = π/6

I = π/12

Problem 5: Evaluate:\int\limits_0^{\pi/2} |sinx - cosx|dx

Solution:

The integral contains mod function.

sin x – cos x = 0

sin x = cos x

tan x = 1

x = π/4

Between limit 0 to π/4, |sinx – cosx| is negative and between π/4 to π/2, |sinx – cosx| is positive

So, we divide the above limits of integral using the following formula

\int\limits_a^b f(x)dx = \int\limits^c_af(x)dx + \int\limits^b_cf(x)dx

\int\limits_0^{\pi/2} |sinx - cosx|dx = \int\limits_{0}^{\pi/4} -(sinx - cosx)dx+\int\limits_{\pi/4}^{\pi/2} (sinx - cosx)dx

\int\limits_0^{\pi/2} |sinx - cosx|dx = \int\limits_{0}^{\pi/4} (cosx - sinx)dx+\int\limits_{\pi/4}^{\pi/2} (sinx - cosx)dx

\int\limits_0^{\pi/2} |sinx - cosx|dx = \big[sinx+cosx\big]_{0}^{\pi/4}+ \big[- cosx-sinx\big]_{\pi/4}^{\pi/2}

\int\limits_0^{\pi/2} |sinx - cosx|dx = \bigg(\frac{2}{\sqrt2}-1\bigg)+\bigg(\frac{2}{\sqrt2}-1\bigg)

\int\limits_0^{\pi/2} |sinx - cosx|dx   = 2√2-2

\int\limits_0^{\pi/2} |sinx - cosx|dx   = 2(√2-1)

Problem 6: Prove that:\int\limits_0^{\pi/2}\text{sin2x log tanx dx}=0

Solution:

Let I =\int\limits_0^{\pi/2}\text{sin2x log tanx dx}   ———(i)

⇒ I =\int\limits_0^{\pi/2}{sin2(\frac{\pi}{2}-x)} \text{log tan}(\frac{\pi}{2}-x) dx

Using property:\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx

⇒ I =\int\limits_0^{\pi/2}\text{sin2x log cotx dx}   ———(ii)

Adding (i) and (ii)

2I =\int\limits_0^{\pi/2}\text{sin2x (log tanx + log cotx)dx}

2I =\int\limits_0^{\pi/2}\text{sin2x (log tanx cotx)dx}

2I =\int\limits_0^{\pi/2}\text{sin2x (log 1)dx}   =0

I = 0

Problem 7: Evaluate:\int\limits_{-\pi}^{\pi}|cosx|dx

Solution:

I =\int\limits_{-\pi}^{\pi}|cosx|dx

|cos x | is an even function

Using property:\int\limits_{-a}^{a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & \text{, if $n$ is even} \\ 0 & \text{, if $n$ is odd} \end{cases}

I = 2\int\limits_{0}^{\pi}|cosx|dx

I = 2 {\int\limits_{0}^{\pi/2}|cosx|dx+\int\limits_{\pi/2}^{\pi}|cosx|dx   }

cos x is negative if π/2< x ≤ π

I = 2{\int\limits_{0}^{\pi/2}cosxdx+\int\limits_{\pi/2}^{\pi}-cosxdx   }

I = 2{\big[sinx\big]_0^{\pi/2}-\big[sinx\big]_{\pi/2}^\pi   }

I = 2 + 2 = 4

Problem 8: Evaluate:\int\limits_1^2\frac{\sqrt x}{\sqrt{3-x}\hspace{0.1cm}+\sqrt{x}}dx

Solution:

I =\int\limits_1^2\frac{\sqrt x}{\sqrt{3-x}\hspace{0.1cm}+\sqrt{x}}dx   ———-(i)

Using property\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a + b - x)dx

I =\int\limits_1^2\frac{\sqrt {3-x}}{\sqrt{3-(3-x)}\hspace{0.1cm}+\sqrt{3-x}}dx

I =\int\limits_1^2\frac{\sqrt {3-x}}{\sqrt{x}\hspace{0.1cm}+\sqrt{3-x}}dx   ———(ii)

Adding (i) and (ii)

2I =\int\limits_1^2\frac{\sqrt {x}+\sqrt {3-x}}{\sqrt{x}\hspace{0.1cm}+\sqrt{3-x}}dx

2I =\int\limits_1^21.dx = \big[x\big]_1^2

2I = 2 – 1

2I = 1

I = 1/2

Problem 9: Show that:\int\limits_0^{\pi/2}f(sin2x)\ sinx\ dx = \sqrt{2}\int\limits_0^{\pi/4}f(cos 2x)\ cosx\ dx

Solution:

I =\int\limits_0^{\pi/2}f(sin2x)\ sinx\ dx

Using property\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx

I =\int\limits_0^{\pi/2}f\{sin 2(\frac{\pi}{2}-x)\}\ sin(\frac{\pi}{2}-x)\ dx

I =\int\limits_0^{\pi/2}f\{sin(\pi-2x)\}\ cosx\ dx

I =\int\limits_0^{\pi/2}f(sin2x)\ cosx\ dx

Adding (i) and (ii)

2I =\int\limits_0^{\pi/2}f(sin2x)\ (sinx+cos x)\ dx

Using property\bold{\int\limits_{0}^{2a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(2a - x) = f(x) \\ 0 & , if f(2a - x) = -f(x)\end{cases}}

2I =2\int\limits_0^{\pi/4}f(sin2x)\ (sinx+cos x)\ dx

2I =2\sqrt{2}\int\limits_0^{\pi/4}f(sin 2x)\big( \frac{1}{\sqrt{2}}sinx + \frac{1}{\sqrt{2}}cosx\big)\ dx

2I =2\sqrt{2}\int\limits_0^{\pi/4}f\{sin 2x\}\ sin(x+\frac{\pi}{4})\ dx

Using property\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{b}f(a+b-x)dx

2I =2\sqrt{2}\int\limits_0^{\pi/4}f\{sin 2(\frac{\pi}{4}-x)\}\ sin(\frac{\pi}{4}-x+\frac{\pi}{4})\ dx

2I =2\sqrt{2}\int\limits_0^{\pi/4}f\{sin (\frac{\pi}{2}-2x)\}\ sin(\frac{\pi}{2}-x)\ dx

2I =2 \sqrt{2}\int\limits_0^{\pi/4}f(cos 2x)\ cosx\ dx

I =\sqrt{2}\int\limits_0^{\pi/4}f(cos 2x)\ cosx\ dx

Problem 10: Evaluate:\int\limits_0^{\pi/2}\frac{x\ sinx\ cosx}{sin^4x+cos^4x}dx

Solution:

I =\int\limits_0^{\pi/2}\frac{x\ sinx\ cosx}{sin^4x+cos^4x}dx

Using property\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx

I =\int\limits_0^{\pi/2}\frac{(\frac{\pi}{2}-x)\ sin(\frac{\pi}{2}-x)\ cos(\frac{\pi}{2}-x)}{sin^4(\frac{\pi}{2}-x)+cos^4(\frac{\pi}{2}-x)}dx

I =\int\limits_0^{\pi/2}\frac{(\frac{\pi}{2}-x)\ cosx\ sinx }{cos^4x+sin^4x}dx

Adding (i) and (ii)

2I =\frac{\pi}{2}\int\limits_0^{\pi/2}\frac{sinx\ cosx}{sin^4x+cos^4x}dx

Let t = sin2x ⇒ dt = 2sinx cosx dx

Limits: x= 0, t = 0 and x = π/2, t =1

2I =\frac{\pi}{4}\int\limits_0^{\pi/2}\frac{1}{(1-t)^2+t^2}dt

2I =\frac{\pi}{8}\int\limits_0^{\pi/2}\frac{1}{\big(t-\frac{1}{2}\big)^2+\big(\frac{1}{2}\big)^2}dt

2I =\frac{\pi}{8}\times2\big[tan^{-1}(2t-1)\big]^1_0

I =\frac{\pi}{8}\big(\frac{\pi}{4}+\frac{\pi}{4}\big)

I =\frac{{\pi}^2}{16}

Also, Check

FAQs on Definite Integrals

Question 1: What are definite integrals?

Answer:

Definite integrals are integrals that are defined under limits i.e., upper and lower limits. It is represented as\bold{\int\limits_{a}^{b}f(x)dx}   where a is the lower limit and b is the upper limit of integration.

Question 2: How are definite integrals evaluated?

Answer:

For evaluating definite integrals following steps are followed:

  • Find the indefinite integral ∫f(x)dx.
  • Evaluate p(a) and p(b) where, p(x) is the antiderivative of f(x), p(a) is the value of antiderivative at x = a, and p(b) is the value of antiderivative at x=b.
  • Calculate p(b) – p(a).
  • The resultant is the desired value of the definite integral.

Question 3: Write the formula for definite integrals.

Answer:

Let p(x) be the antiderivative of function f(x) defined in the interval [a, b] then, the definite integral is given by:

\bold{\int\limits_{a}^{b}f(x)dx}   = p(b) – p(a)

Question 4: What does the result of definite integral represent?

Answer:

The result obtained by solving the definite integral represents the area under the curve that is evaluated.


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