Integrals Class 12 Notes Maths Chapter 7 - CBSE


What are Integrals ?

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    The process of integration is an opposite process to
    that of differentiation.

    ∫ f (x) dx = F(x) + C
    Where ∫ —› symbol of integration
    f(x) —› integrand
    dx —› element of integration
    F —› primitive
    C —› constant of integration In differential calculus, we are given a function and we are required to differentiate it, but in integral calculus we required to find a function whose differential coefficient (or derivative) is given.
    Hence if,

    d(F(x) + c) = f(x)/dx

    We write that, ∫ f(x)dx = F(x) + c

    Standard Integrals

    $$\text{(i)\space }\int 0\space \text{dx}=\text{C(constant)}\\ \text{(ii)}\int 1\text{dx} = x+\text{C}\\ \text{(iii)}\int x^n\text{dx}= \frac{x^{n+1}}{n+1}+\text{C} \space\text{if} ≠ -1\\ \text{(iv)} \int\frac{1}{x}\text{dx}=\text{log}_e\text{x}+\text{C}\\ \text{(v)}\int e^x\text{dx}=e^x+\text{C}$$

    $$\text{(vi)}\int a^x\text{dx}=\frac{a^x}{\text{log}_ea}+\text{C}\\ \text{(vii)} \int \text{sin}\space x \space \text{dx\space=\space -cos}\space x+\text{C}\\ \text{(viii)}\int \text{cos}\space x \space\text{dx}=\text{sin}\space x+\text{C}\\ \text{(ix)}\int \text{sec}^2x\space\text{dx}= \text{tan}\space x+\text{C}\\ \text{(x)}\int\text{cosec}^2x \space \text{dx} =-\text{cot}\space x+\text{C}$$

    $$\text{(xi)}\int \text{sec}\space x .\text{tan}\space x\space \text{dx}=\text{sec}\space x+\text{C}\\ \text{(xii)} \int \text{cosec}\space x\space.\space \text{cot}\space x\space\text{dx} =-\text{cosec x}+\text{C}\\\text{(xiii)}\int \text{cot}\space x\space\text{dx} = \text{log}|\text{sin} \space x|+\text{C}\\ \text{(xiv)}\int\text{tan}\space x \space\text{dx}=-\text{log}|\text{cos}\space x|+\text{C}\\\text{(xv)}\int\text{sec}\space x\space\text{dx}= \text{log}|\text{sec}\space x + \text{tan \space x}|+\text{C}=\text{log}|\text{tan}(\frac{\pi}{2}+\frac{x}{2})|+ \text{C}\\ \text{(xvi)} \int \text{cosec}\space x\space\text{dx}=\text{log}|\text{cosec}\space x-\text{cot}\space x|=\text{log}|\text{tan} \frac{x}{2}|+\text{C}$$


    Methods Of Integration

    We need to develop additional techniques or methods for finding the integrals by reducing them into standard forms. Here are some important methods:
    A. Integration by Substitution
    B. Integration by Partial Fraction
    C. Integration by Parts

    A. Integration by Substitution

    The given integral ∫f(x)dx can be transformed into another form by changing independent variables 'x' to 't' by substituting x=g(t). Consider I = ∫ f(x)dx Put x=g(t) so that dx/dt= g'(t) ==> dx=g'(t). dt → I= ∫ f(x)dx = ∫ f{g(t)}. g'(t)dt There are some useful substitution: (a) (i) If the integrand contains a t-ratio of f(x)or logarithm of f(x) or an exponential function in which the index is f(x), put f(x)=t. (ii) If the integrand is a rational function of ex, put ex = t. (b) For evaluating integrals ∫ sinnx.dx or ∫ cosnx.dx, where n = positive integer, (i) Put sin x = t, if the index of cos x is an odd positive integer. (ii) Put cos x = t, if the index of sin x is an odd positive integer. (iii)Express sinn x or cosn x in terms of cosines of multiples of angles by using 2sin2 x = 1-cos 2x or 2cos2 x = 1 + cos 2x if n is an even positive integer. (c) For evaluating integrals ∫ secn x dx or ∫ cosecn x dx where n = positive integer, (i) Put tan x = t, if the index of sec x is a positive even integer. (ii) Put cot x = t, if the index of cosec x is a positive even integer.

    $$\text{(d)}\space \text{(i)}\space\text{If}\space \sqrt{a^2-x^2}\space\text{occurs in the integrand, put x = a sin}\space \alpha.\\ \qquad \text{(ii)}\space\text{If}\sqrt{a^2+x^2}\space\text{occurs in the integrand, put x = a tan} \space \alpha.\\ \qquad\text{(iii) If}\sqrt{x^2-a^2}\space\text{occurs in the integrand, put x = a sec}\space \alpha.$$

    B. Partial fractions

    Let f(x) =g(x)/h(x) be a proper rational function. First of all, we split up the denominator h(x) as the product of non-repeated linear or repeated linear factors or non-repeated or repeated quadratics (which cannot be split into real linear factors). Then f(x) can be written as the sum of the fractions in which the numerator is either a constant or real linear polynomials. For this the following points should be kept in mind.

    (a) Corresponding to each non-repeated linear factor a x+ b, there is a partial fraction of the form A/(ax+b)

    (b) Corresponding to each repeated linear factor (ax+b)2 there is sum of two terms of the form A/(ax+b)+B/(ax+b)2
    (c) Corresponding to each non-repeated quadratic factor ax2+bx+c (which cannot be put as the product of linear factors) there is a partial factor of the form (ax+b)/ax2+bx+c
    (d) Corresponding to each repeated quadratic factor (ax2+bx+c)2 (which cannot be put as the product of linear factors), there is a sum of two terms of the form (Ax+B)/(ax2+bx+c) +(Cx + D)/(ax2 + bx + c)2
    (e) When only even powers of x occur both in the numerator and denominator put x2 = y the resulting algebraic function in y into partial fraction and then put y = x2 in the resulting partial fraction.

    C. Integration By Parts

    Consider u and v as two functions of x, then integration
    by parts says that; ∫ u.v dx = u { ∫ v dx} - ∫{du/dx. ∫ v dx } dx Where, u is known as first function and
    v is known as second function.
    v is known as second function.
    i.e., Integral of product of two = (first function *
    (integral of second function) - integral of
    {(differentiation of first function) * (integral of
    second function)}
    We decide the function as first which comes first in
    order from top to bottom.
    INVERSE —› I
    Or which comes first in word 'ILATE'.

    Fundamental Theorem Of Calculus

    The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function. The two operations are inverses of each. other apart from a constant value which depends where one starts to compute area. The first part of the theorem, called first fundamental
    theorem of calculus. Let f be a continuous real-valued function defined on a closed interval [a,b] and the function F(x) is defined by

    $$\text{F(x)} = \int^x_a \text{f(t)}\space \text{dt}$$

    then F’(x) = f(x)
    The second part of the theorem called second fundamental theorem of calculus. Let f be a continuous real-valued function on a closed interval [a,b], then

    $$ \int^b_a \text{f(x)}\text{dx}= \text{F(b)}-\text{F(a)}$$

    Definite Integral

    The definite integral has a unique value. A definite integral is denoted by

    $$ \int^b_a \text{f(x)}\text{dx},$$

    where a is called the lower limit of the integral and b is called the upper limit of the integral.

    Standard Integrals

    $$\text{(i)} \int^b_a \text{f(x)}\text{dx}=\text{F(b)\space-\space F(a)}\space \text{where}\int \text{f(x)dx=\space \text{F(x)\space +\space\text{C}}}\\\text{(ii)} \int^b_a\text{f(x)dx}=\int^b_a\text{f(t)}\text{dt}\\\text{(iii)}\int^b_a \text{f(x)}\text{dx}=-\int^a_b \text{f(x)dx}\\\text{(iv)}\int^b_a\text{f(x)dx} = \int^c_a\text{f(x)dx}\space+\space \int^b_c\text{f(x)dx},\\ \text{where the point c lies between a and b}\\\text{(v)}\int^b_a\text{f(x)dx}=\int^b_a\text{f(a\space+\space b-x)dx}\\\text{(vi)}\int^a_0 \text{f(x)dx}=\int^a_0 \text{f(a-x)dx}\\\text{(vii)}\int^\text{2a}_0\text{f(x)dx}=\int^a_0\text{{f(x)dx + f(2a-x)}}\text{dx}\\\text{(viii)} \int^{2a}_0\text{f(x)dx}= 2\int^a_0\text{f(x)dx,}\space \text{if}\space \text{f(2a-x)}= \text{f(x)}\space \text{and} \space 0, \text{if} \space \text{f(2a-x)}=-\text{f(x)}\\ \text{(ix)}\int^a_{-a} \text{f(x)dx}=2\int^a_0 \text{f(x)dx,}\space \text{if} \space \text{f(-x)}=\text{f(x)} \space \text{and} \space 0, \text{if f(-x) = -f(x)}$$

    Evaluation Of Definite Integrals By Substitution

    $$\text{To evaluate}\int^b_a \text{f(x)dx,} \space \text{by substitution, the following steps are useful}$$

    • Consider the integral without limits and substitute, y=f(x) or x=g(y) to reduce the given integral to a known form.
    • lntegrate the new integrand with respect to the new variable without mentioning the constant of integration.
    • Resubstitute for the new variables and write the answer in terms of the original variable.
    • Find the values of answer obtained in III step at he given limits of integral and find the difference of the values at the upper and lower limits.