If F 3 21 F Â² is Continuous and Â«93f Â² X dx 10 Then the Value of F 9 is

ELet p(x) 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 integration is not included in the evaluation of the definite integral.
$\int\limits_{a}^{b}f(x)dx$ is read as "integral of f(x) from a to b"

Steps to find 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:

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

Definite Integrals by Substitution

For the integral $\int_{a}^{b}f(g(x))g'(x)dx$ . Let g(x) = t, then g'(x) dx = dt where for x = a , t = g(a) and for x = b, t = g(b).

$\int\limits_{a}^{b}f(g(x))g'(x)dx = \int\limits_{g(a)}^{g(b)}f(t)dt$

If the variable is changed in the definite integral then substitution of a new variable affects the integrand, the differential (i.e. dx), and the limits.

The limits of the new variable t are the values of t corresponding to the values of the original variable x. It can be obtained by putting values of x in the substitution relation of x and t.

Properties of Definite Integral

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

Proof:

Let p(x) be a 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)

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 the 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

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

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

⇒ $\int\limits_{0}^{a}f(x)dx =\int\limits_{0}^{a}f(a - x)dx$ [ By 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 $\int\limits_{a}^{2a}f(x)dx$

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]

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

Solved Example on Definite Integrals

Problem 1: Evaluate:

(i) $\bold{\int\limits_{1}^{2}x^2 dx}$

(ii) $\bold{\int\limits_{0}^{1}\frac{1}{(2x - 3)} dx}$

(iii) $\bold{\int\limits_{0}^{\pi/4}tan^2x dx}$

Solution:

(i) $\int\limits_{1}^{2}x^2dx$ = $\big[x^3\big]_{1}^{2}$

= [2³ – 1³]

= 8 – 1

$\int\limits_{1}^{2}x^2$ dx = 7

(ii) $\int\limits_{0}^{1}\frac{1}{(2x - 3)} dx$ = $\frac{1}{2}\big[log(2x-3)\big]_{0}^{1}$

= (1/2)[log|-1| – log|-3| ]

= (1/2)[ log 1 – log 3]

= (1/2)[0 – log 3]

$\int\limits_{0}^{1}\frac{1}{(2x-3)} dx$ = (1/2)log 3

(iii) $\int\limits_{0}^{\pi/4}tan^2x dx$ = $\int\limits_{0}^{\pi/4}$ (sec² x – 1) dx

= $\big[tanx - x\big]_{0}^{\pi/4}$

= [tan(π/4) – (π/4)] – [tan 0 – 0 ]

$\int\limits_{0}^{\pi/4}tan^2x dx$ = 1 – (π/4)

Problem 2: Evaluate: $\bold{\int\limits_{0}^{1}\frac{2x}{5x^2 +1}}$

Solution:

Let 5x²+ 1 = t. Then, d(5x² + 1) = dt ⇒ 10 x dx = dt

For limits : Lower limit ⇒ x = 0 then t = 5x²+1 = 1 and Upper limit ⇒ x = 1 then t = 5x² + 1 = 6

$\int\limits_{0}^{1}\frac{2x}{5x^2 +1}=\int\limits_{1}^{6}\frac{2x}{t}.\frac{dt}{10x}$

= $\frac{1}{5}\int\limits_{1}^{6}\frac{1}{t}dt$

= $\frac{1}{5}\big[log\hspace{0.1cm}t\big]_1^6$

= (1/5) [log 6 – log 1]

$\int\limits_{0}^{1}\frac{2x}{5x^2 +1}$ = (1/5) log 6

Problem 3: Evaluate : $\int\limits_{-1}^{1} \text {f(x)dx , where f(x) = }\begin{cases} 1 - 2x , x \le 0\\ 1 + 2x , x\ge 0\end{cases}$

Solution:

$f(x) = \begin{cases} 1 - 2x \hspace{0.2cm}, x \le 0\\ 1 + 2x \hspace{0.2cm}, x\ge 0\end{cases}$

$\int\limits_{-1}^{1} f(x) dx = \int\limits_{-1}^0 f(x)dx \hspace{0.2cm}+ \int\limits_{0}^1 f(x)dx$

$\int\limits_{-1}^{1} f(x) dx = \int\limits_{-1}^0 (1-2x)dx \hspace{0.2cm}+ \int\limits_{0}^1 (1 + 2x)dx$ [Using definition of f(x)]

$\int\limits_{-1}^{1} f(x) dx = \big[x-x^2\big]_{-1}^0 \hspace{0.2cm}+ \big[x + x^2\big]_{0}^1$

= [0 – ( -1 – 1)] + [(1 + 1) – (0)]

$\int\limits_{-1}^{1} f(x) dx = 4$

Problem 4: Evaluate: $\int\limits_0^\pi|cos x| dx$

Solution:

$|cos x| =\begin{cases} cosx\hspace{0.4cm}, 0\le x\le \pi/2\\ -cosx \hspace{0.2cm}, \pi/2\le x\le \pi\end{cases}$

$\int\limits_0^\pi|cos x| dx = \int\limits_0^{\pi/2}|cos x| dx + \int\limits_{\pi/2}^\pi|cos x| dx$

⇒ $\int\limits_0^\pi|cos x| dx = \int\limits_0^{\pi/2}|cos x| dx + \int\limits_{\pi/2}^\pi (-cos x) dx$

⇒ $\int\limits_0^\pi|cos x| dx = \big[cos x\big]_0^{\pi/2} - [sin x\big] _{\pi/2}^\pi$

= 1 + 1

$\int\limits_0^\pi|cos x| dx = 2$

Problems 5: Evaluate: $\int\limits_0^{\pi/2}log tan\hspace{0.1cm}x\hspace{0.1cm}dx$

Solution:

I = $\int\limits_0^{\pi/2}log tan\hspace{0.1cm}x\hspace{0.1cm}dx$ ———————(i)

I = $\int\limits_0^{\pi/2}log tan\hspace{0.1cm}(\frac{\pi}{2}-x)\hspace{0.1cm}dx$

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

I = $\int\limits_0^{\pi/2}log \hspace{0.1cm}cotx\hspace{0.1cm}dx$ ——————-(ii)

Adding (i) and (ii)

2I = $\int\limits_0^{\pi/2}[log \hspace{0.1cm}tanx\hspace{0.1cm} +log \hspace{0.1cm}cotx\hspace{0.1cm}]dx$

2I = $\int\limits_0^{\pi/2}[log \hspace{0.1cm}(tanx\hspace{0.1cm} cotx)\hspace{0.1cm}]dx$

2I = $\int\limits_0^{\pi/2}log1.dx$

2I = $\int\limits_0^{\pi/2}0.dx$

I = 0

Problem 6 : 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

FAQs on Definite Integrals

Question 1: What is meant by definite integrals?

Answer:

Definite integrals are integrals that are defined under proper limits i.e. their upper and lower limits are specified. It is represented as ∫_b ^a f(x) dx where a is the upper limit and b is the lower limit of integration.

Question 2: How are definite integrals simplified?

Answer:

For simplifying definite integrals use the following steps:

Simplify the integral normally.

Substitute upper and lower limits to the answer of integration.

Subtract both the answer obtained in step 2

Question 3: Write the formula for solving definite integrals.

Answer:

Suppose a definite integral of a function f(x) in the interval [a, b] is required, then,

∫_b ^a f(x) dx = F(a) – F(b)

where, ∫ f(x) dx = F(x) + C

Question 4: What does the value obtained from solving definite integral represent? Can it be negative?

Answer:

The value obtained from solving the definite integral represents the area. Yes, it can also be negative.

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Source: https://www.geeksforgeeks.org/calculate-definite-integral/

If F 3 21 F Â² is Continuous and Â«93f Â² X dx 10 Then the Value of F 9 is

Steps to find Definite Integrals

Definite Integrals by Substitution

Properties of Definite Integral

Solved Example on Definite Integrals

FAQs on Definite Integrals

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