Definite Integration

The definite integral has a unique value. A definite integral is denoted by ${\int }_a^{b}f(x)dx$, where a is called the lower limit of the integral and b is called the upper limit of the integral.

1.0Definite Integration Definition

A definite integral is denoted by ${\int }_a^{b}f(x)dx$ which represents the algebraic area bounded by the curve y = f(x), the ordinates x = a, x = b and the x-axis.

2.0Fundamental Theorem of Calculus

P–1 : If f is continuous on [a, b] then the function g defined by g(x) = ${\int }_a^{x}f(t)dt$, a ≤ x ≤ b is continuous on [a, b] and differentiable on (a, b) and g'(x) = f(x)

P–1 : If f is continuous on [a, b] then ${\int }_a^{b}f(x)dx$ = F(b) – F(a) where F is any antiderivative of F, such that F' = f

3.0Properties of Definite Integrals

P–1 : ${\int }_a^{b}f(x)dx={\int }_a^{b}f(t)dt$ (change of variable does not change value of integral)

P–2 : ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$

P–3 :  ${\int }_a^{b}f(x)={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$

P–4 : ${\int }_{-a}^{a}f(x)dx={\int }_0^a(f(x)+f(-x))dx=[\begin{array}{ll}0& \text{ if }f(x)\text{ is odd }\\ 2{\int }_0^{a}f(x)dx& \text{ if }f(x)\text{ is even }\end{array}$

P–5: ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx\text{ or }{\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$ (King)

P–6: ${\int }_0^{2a}f(x)dx={\int }_0^{a}f(x)dx+{\int }_0^{a}f(2a-x)dx=$

$[\begin{array}{cc}0& \text{ if }f(2a-x)=-f(x)\\ 2{\int }_0^{a}f(x)dx& \text{ if }f(2a-x)=f(x)\end{array}$(Queen)

4.0Walli’s Theorem

(a) ${\int }_0^{\pi /2}{\sin }^{n}xdx={\int }_0^{\pi /2}{\cos }^{n}xdx=\frac{(n-1)(n-3)\dots ..(1\text{ or }2)}{n(n-2)\dots ..(1\text{ or }2)}K$

$\text{ where }K=\{\begin{array}{ll}\pi /2& \text{ if }n\text{ is even }\\ 1& \text{ if }n\text{ is odd }\end{array}$

(b) ${\int }_0^{\pi /2}{\sin }^{n}x\cdot {\cos }^{m}xdx=\frac{[(n-1)(n-3)(n-5)\dots .1\text{ or }2][(m-1)(m-3)\dots .\dots \text{ or }2]}{(m+n)(m+n-2)(m+n-4)\dots .1\text{ or }2}K\text{ Where }K=\{\begin{array}{ll}\frac{\pi }{2}& \text{ if both }m\text{ and }n\text{ are even }(m,n\in N)\\ 1& \text{ otherwise }\end{array}$

5.0Derivative of Antiderivatives(Newton-Leibnitz Theorem)

$\text{ If }F(x)={\int }_{g(x)}^{h(x)}f(t)dt\text{, then }\frac{dF(x)}{dx}=h^{\prime }(x)f(h(x))-g^{\prime }(x)f(g(x))$

$\text{ Proof: Let }P(t)=\int f(t)dt\Rightarrow F(x)={\int }_{g(x)}^{h(x)}f(t)dt=P(h(x))-P(g(x))$

$\Rightarrow \frac{dF(x)}{dx}=P^{\prime }(h(x))h^{\prime }(x)-P^{\prime }(g(x))g^{\prime }(x)=f(h(x))h^{\prime }(x)-f(g(x))g^{\prime }(x)$

$\text{ If }F(x)=\text{, then }=h^{\prime }(x)f(h(x))-g^{\prime }(x)f(g(x))$

$\text{ Proof : Let }P(t)=\int f(t)dt\Rightarrow F(x)==P(h(x))-P(g(x))$

$\Rightarrow \frac{dF(x)}{dx}=P^{\prime }(h(x))h^{\prime }(x)-P^{\prime }(g(x))g^{\prime }(x)=f(h(x))h^{\prime }(x)-f(g(x))g^{\prime }(x)$

6.0Definite Integration as the limit of Sum

A definite integral can be evaluated using Riemann sums, which approximate the area under the curve by summing the areas of rectangles

${\int }_a^{b}f(x)dx={\lim }_{n\to \infty }{\sum }_{i=1}^{n}f\left(x_i\right)\Delta x$

where

$\Delta \mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}$

and x_i is a sample point in the i^th subinterval.

7.0Estimation of Definite Integral and General Inequality

1. f(x) ≤ g(x) ≤ h(x) in [a, b] then ${\int }_a^{b}f(x)dx\le {\int }_a^{b}g(x)dx\le {\int }_a^{b}h(x)dx$
2. If M and M are respectively the least and greatest value of f(x) in [a, b] then m (b – a) ≤ ${\int }_a^{b}f(x)dx\le M(b-a)$

8.0Definite Integration Questions

Example 1: ${\int }_0^{\ln 2}\frac{e^x}{1+e^x}dx$

Solution:

$1+e^x=t\mathrm{x}=0\Rightarrow \mathrm{t}=2$

$e^{x}dx=dtx=\ln 2\Rightarrow \mathrm{t}=1+2=3$

${\int }_2^3\frac{dt}{t}=(\ln |t|{)}_2^3=\ln \left(\frac{3}{2}\right)$

Example 2: ${\int }_{\alpha }^{\beta }\frac{dx}{\sqrt{(x-\alpha )(\beta -x)}}\beta >\alpha$

Solution:

Put $\sqrt{x-\alpha }=t$

$\int \frac{2tdt}{t\sqrt{(\beta -\left(t^2+\alpha \right)}}\Rightarrow 2\int \frac{dt}{\sqrt{(\beta -\alpha )-t^2}}$

$\Rightarrow 2{\sin }^{-1}\left(\frac{t}{\sqrt{\beta -\alpha }}\right)\Rightarrow {\left[2{\sin }^{-1}\left(\frac{\sqrt{x-\alpha }}{\sqrt{\beta -\alpha }}\right)\right]}_{\alpha }^{\beta }$

$\Rightarrow 2\left(\frac{\pi }{2}-0\right)\Rightarrow \pi$

Example 3: ${\int }_{-\pi /2}^{\pi /2}{\sin }^{4}x{\cos }^{6}xdx=$

(A) $\frac{3\pi }{64}$

(B) $\frac{3\pi }{572}$

(C) $\frac{3\pi }{256}$

(D) $\frac{3\pi }{128}$

Ans. (C)

Solution:

$\mathrm{I}={\int }_{-\pi /2}^{\pi /2}{\sin }^{4}x{\cos }^{6}xdx=2{\int }_0^{\pi /2}{\sin }^{4}x{\cos }^{6}x\cdot dx=2\frac{(3.1)(5\cdot 3\cdot 1)}{10.8\cdot 6\cdot 4\cdot 2}\cdot \frac{\pi }{2}=\frac{3\pi }{256}$

Example 4: ${\int }_0^1\frac{{\tan }^{-1}x}{x}dx=\lambda {\int }_0^{\pi /2}\frac{\theta }{\sin \theta }d\theta$, then find λ.

Ans. 1/2

Solution:

$\text{ L.H.S. }\int \frac{{\tan }^{-1}x}{x}dx$

Put

$={\int }_0^{\pi /4}\frac{\mathrm{t}}{\tan t}\cdot {\sec }^{2}tdt={\int }_0^{\pi /4}\frac{\mathrm{t}}{\sin t\cos t}dt={\int }_0^{\pi /4}\frac{2\mathrm{t}}{\sin 2t}dt$

2t = θ

2dt= dθ

$=\frac{1}{2}{\int }_0^{\pi /2}\frac{\theta }{\sin \theta }d\theta$

∴$\lambda =\frac{1}{2}$                                                                                                                                                                                                                                                                                                  

Example 5: Evaluate ${\int }_2^8|x-5|dx$.

Solution:

${\int }_2^8|x-5|dx={\int }_2^5(-x+5)dx+{\int }_5^8(x-5)dx=9$

Example 6: $\text{ Evaluate }{\int }_{-1}^1\frac{3^x+3^{-x}}{1+3^x}dx$

Solution:

${\int }_{-1}^1\frac{3^x+3^{-x}}{1+3^x}dx={\int }_0^1\left(\frac{3^x+3^{-x}}{1+3^x}+\frac{3^{-x}+3^x}{1+3^{-x}}\right)dx={\int }_0^1\left(\frac{3^x+3^{-x}}{1+3^x}+\frac{3^x\left(3^{-x}+3^x\right)}{1+3^x}\right)$

$={\int }_0^1\left(3^x+3^{-x}\right)dx={\left(\frac{3^x}{\ln 3}-\frac{3^{-x}}{\ln 3}\right)}_0^1=\left(\frac{3}{\ln 3}-\frac{3^{-1}}{\ln 3}\right)-\left(\frac{1}{\ln 3}-\frac{1}{\ln 3}\right)=\frac{1}{\ln 3}\left[3-\frac{1}{3}\right]=\frac{8}{3\ln 3}$

Example 7: $\text{ Prove that }{\int }_0^{\pi /2}\log (\sin x)dx={\int }_0^{\pi /2}\log (\cos x)dx=-\frac{\pi }{2}\log 2$

Solution:

Let $I={\int }_0^{\pi /2}\log (\sin x)dx$ ...(i)

then $I={\int }_0^{\pi /2}\log \sin \left(\frac{\pi }{2}-x\right)dx={\int }_0^{\pi /2}\log (\cos x)dx$ ...(ii)

adding (i) and (ii), we get

$2I={\int }_0^{\pi /2}\log \sin xdx+{\int }_0^{\pi /2}\log \cos xdx={\int }_0^{\pi /2}(\log \sin x+\log \cos x)dx$

$\Rightarrow 2I={\int }_0^{\pi /2}\log (\sin x\cos x)dx={\int }_0^{\pi /2}\log \left(\frac{2\sin x\cos x}{2}\right)dx$

$={\int }_0^{\pi /2}\log \left(\frac{\sin 2x}{2}\right)dx={\int }_0^{\pi /2}\log (\sin 2x)dx-{\int }_0^{\pi /2}(\log 2)dx={\int }_0^{\pi /2}\log \sin 2x\cdot dx-(\log 2)(x{)}_0^{\pi /2}$

$\Rightarrow 2I={\int }_0^{\pi /2}\log (\sin 2x)dx-\frac{\pi }{2}\log 2$ ...(iii)

Let

$I_1={\int }_0^{\pi /2}\log (\sin 2x)dx$,

putting 2 x=t, we get

${\mathrm{I}}_1={\int }_0^{\pi }\log (\sin t)\frac{dt}{2}=\frac{1}{2}{\int }_0^{\pi }\log (\sin t)dt=\frac{1}{2}\cdot 2{\int }_0^{\pi /2}\log (\sin t)dt$

${\mathrm{I}}_1={\int }_0^{\pi /2}\log (\sin x)dx$

∴ (iii) becomes; $2I=I-\frac{\pi }{2}\log 2$

Hence ${\int }_0^{\pi /2}\log \sin xdx=-\frac{\pi }{2}\log 2$

Example 8: Find the slope of the tangent to the curve $y={\int }_x^{x^2}{\cos }^{-1}{t}^{2}dt\text{ at }x=\frac{1}{\sqrt[4]{2}}$

Solution:

Given curve is

$y={\int }_x^{x^2}{\cos }^{-1}{t}^{2}dt;\frac{dy}{dx}=\frac{d}{dx}{\int }_x^{x^2}{\cos }^{-1}{t}^{2}dt$

using Leibnitz theorem,  $\frac{dy}{dx}=2x{\cos }^{-1}{x}^4-{\cos }^{-1}{x}^2$

${\left(\frac{dy}{dx}\right)}_{x=\frac{1}{\sqrt[4]{2}}}=\frac{2}{2^{1/4}}{\cos }^{-1}\frac{1}{2}-{\cos }^{-1}\frac{1}{\sqrt{2}}=2^{3/4}\frac{\pi }{3}-\frac{\pi }{4}=\left(\frac{\sqrt[4]{8}}{3}-\frac{1}{4}\right)\pi$

9.0Practice Problems on Definite Integration

a. If $f(x)=|x|+|x-1|+|x-2|,x\in R\text{ then }{\int }_0^{3}f(x)\mathrm{dx}=$

b. Evaluate: ${\int }_0^1\frac{x^2-1}{1+x^2+x^4}dx$

c. Evaluate: ${\int }_0^{\pi /2}\frac{dx}{1+\cos \theta \cdot \cos x}\theta \in (0,\pi )$

d. The value of the integral ${\int }_{-1}^1\log \left(x+\sqrt{x^2+1}\right)dx$ is:

e. Evaluate: ${\int }_{-1}^1\frac{\left(2x^{332}+x^{998}+4x^{1668}\cdot \sin x^{691}\right)}{1+x^{666}}dx$

Answers:

a. $\frac{19}{2}$

b. $\frac{1}{2}\ln \left(\frac{1}{3}\right)$

c. $\frac{\theta }{\sin \theta }$

d. 0

e. $\frac{2}{333}\left[\frac{\pi }{4}+1\right]=\frac{\pi +4}{666}$

10.0Solved Questions on Definite Integration

1. How do you define a definite integral?

Ans: A definite integral is defined as the integral of a function f(x) from a to b, denoted as ${\int }_a^{b}f(x)dx$, where 'a' and 'b' are the limits of integration.

2. How do you perform definite integration by parts?

Ans: Integration by parts for definite integrals is given by: ${\int }_a^{b}udv={uv|}_a^b-{\int }_a^{b}vdu$ where u and v are differentiable functions.

3. How do you evaluate a definite integral using the limit of a sum?

Ans: A definite integral can be evaluated using Riemann sums, which approximate the area under the curve by summing the areas of rectangles:

${\int }_a^{b}f(x)dx={\lim }_{n\to \infty }{\sum }_{i=1}^{n}f\left(x_i^{\ast }\right)\Delta x$ , where  $\Delta x=\frac{b-a}{n}\text{ and }{x}_i^{\ast }$ is a sample point in the i^th subinterval.