Mathematics Fundamental Theorem of Calculus For CBSE-NCERT
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color{red}♦ Fundamental Theorem of Calculus

### Fundamental Theorem of Calculus

\color{green} ✍️ color{blue} text{Area function}

=> We have defined int_a^bf(x) dx as the area of the region bounded by the curve y = f(x) the ordinates x = a and x = b and x-axis.

=> Let x be a given point in [a, b]. Then int_a^x f(x) dx represents the area of the shaded region in Fig.

=> In other words, the area of this shaded region is a function of x. We denote this function of x by A(x).

color {brown} {A(x) = int_a^x f(x) dx} .........................(1)

\color{green} ✍️ color{blue} text{ First fundamental theorem of integral calculus}

Theorem 1 : Let f be a continuous function on the closed interval [a, b] and let A (x) be the area function.

Then color{green}{A′(x) = f (x),} for all x ∈ [a, b].

\color{green} ✍️ color{blue} text{ Second fundamental theorem of integral calculus}

We state below an important theorem which enables us to evaluate definite integrals by making use of anti derivative.

Theorem 2 : Let f be continuous function defined on the closed interval [a, b] and F be an anti derivative of f.

Then, color{brown} {int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)}

color{green} text{Remarks}

(i) Theorem 2 tells us that int_a^b f(x) dx = (value of the anti derivative F of f at the upper limit b – value of the same anti derivative at the lower limit a).

(iii) In int_a^b f(x) dx , the function f needs to be well defined and continuous in [a, b]. For instance, the consideration of definite integral int_(-2)^3 x (x^2 -1)^(1/2) dx is erroneous since the function f expressed by f (x) = x(x^2 -1)^(1/2) is not defined in a portion -1 < x < 1  of the closed interval [– 2, 3].

text{Steps for calculating} color{brown} {int_a^bf(x)dx}

(i) Find the indefinite integral ∫ f (x) dx . Let this be F(x) There is no need to keep integration constant C
because if we consider F(x) + C instead of F(x), we get

=>color{brown}{ int_a^b f(x) dx = [F(x) + C]_a^b = [F(b) +C ] - [F(a) +C] = F(b) - F(a)}

Thus, the arbitrary constant disappears in evaluating the value of the definite integral.

(ii) Evaluate F(b) – F(a) = [F ( x)]_a^b , which is the value of int_a^b f(x) dx
Q 3175480366

Evaluate the following integrals:

(i)  ∫_(2)^3 x^2 dx

(ii) ∫_(4)^9 (sqrt x)/ ((30 -x^(3/2))^2 dx

(iii)  ∫_(1)^2 (x dx)/( (x+1) ( x+2) )

(iv) ∫_(0)^(pi/4) sin^3 2t cos 2 t dt
Class 12 Chapter 7 Example 27
Solution:

(i) Let I = ∫_(2)^3 x^2 dx . Since  ∫ x^2 dx = x^3/3 = F(x)

Therefore, by the second fundamental theorem, we get

I=F(3) - F(2) = 27/3 - 8/3 = 19/3

(ii) Let I = ∫_(4)^9 (sqrt x)/( (30- x^(3/2))^2) dx . We first find the anti derivative of the integrand.

Put 30- x^3/2 = t. Then -3/2 sqrt x dx = dt or sqrt x dx = -2/3 dt

Thus,  ∫ (sqrt x)/( (30- x^(3/2))^2 ) dx = -2/3 ∫ (dt)/(t^2) = 2/3 [1/t] = 2/3 [1/(30- x^(3/2)) ] =F (x)

Therefore, by the second fundamental theorem of calculus, we have

I= F (9) -F(4) =2/3 [ 1/(30 - x^(3/2)) ]_(4)^9

 = 2/3 [1/(30-27) - 1/(30-8)] = 2/3 [1/3 - 1/22] = 19/99

(iii) Let I = ∫_(1)^2 (x dx )/( (x+1) (x+2) )

Using partial fraction, we get x/( (x+1) (x+2) ) = (-1)/(x+1) +2/(x+2)

So  ∫ (x dx)/( (x+1) (x+2) ) = - log | x+1 | + 2 log | x+2 | = F(x)

Therefore, by the second fundamental theorem of calculus, we have

I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3]

 = -3 log 3 + log 2 + 2 log 4 = log (32/27)

(iv) Let I = ∫_(0)^(pi/4) sin^3 2t cos 2 t dt Consider  ∫ sin^3 2t cos 2 t dt

Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = 1/2 du

So  ∫ sin^3 2t cos 2t dt = 1/2 ∫ u^3 du

= 1/8 [ u^4 ] = 1/8 sin^4 2t = F (t ) say

Therefore, by the second fundamental theorem of integral calculus

I= F (pi/4) -F (0) = 1/8 [ sin^4 (pi/2) - sin^4 0 ] = 1/8