Notes On Comparison Between Differentiation and Integration - CBSE Class 12 Maths
1. Differentiation and integration operate on functions. d/dx[p1f1(x) + p2f2(x)] = p1 d/dx f1(x) + p2 d/dx f2(x) ∫[p1f1(x) + p2f2(x)] dx = p1 ∫ f1(x) dx + p2 ∫ f2(x) dx 2. Differentiation and integration satisfy the property of linearity. a) Property of linearity in differentiation: d/dx[p1f1(x) + p2f2(x)] = p1 d/dx f1(x) + p2 d/dx f2(x) b) Property of linearity in integration: ∫[p1f1(x) + p2f2(x)] dx = p1 ∫ f1(x) dx + p2 ∫ f2(x) dx Let y = 10 log x + 4 sin x d/dx(y) = d/dx (10 log x + 4 sin x) = 10 d/dx (log x) + 4 d/dx (sin x) ∫[10/x + 4cos x] dx = 10 ∫1/x dx + 4 ∫ cos x dx Integrable function: If a function f(x) has an integral, then f(x) is called an integrable function. 3. All functions may not be differentiable, and all functions may not be integrable. Example: Prove that the function f(x) = |x - 2|, x ∈ R is not differentiable at x = 2 Given function is f(x) = |x - 2|, x ∈ R $\text{}\mathrm{| x-2 |}=\left\{\begin{array}{ll}x-2& \hfill \phantom{\rule{10}{0ex}}\mathrm{when x}\ge 2\\ -\left(x-2\right)& \hfill \phantom{\rule{10}{0ex}}\mathrm{when x}<2\end{array}\right\$ and LHL =  $\underset{h\to 0⁻}{\text{lim}}\frac{\text{f(2+h) - f(2)}}{\text{h}}$   = (-(2+ h - 2) - 0)/h = -h/h = -1 RHL = =  $\underset{h\to 0⁻}{\text{lim}}\frac{\text{f(2+h) - f(2)}}{\text{h}}$   = ((2 + h + 2) - 0)/h = h/h = 1 LHL ≠ RHL Hence,=    does not exist. Therefore, f is not differentiable at 2. Ex: Let f(x) = sinx / x ∫f(x)dx = ∫sinx/x dx d/dx(?) = sin x /x Hence, f(x) = sin x /x is not integrable. 4. The derivative of a function, if it exists, is unique. The integral of a function, if it exists, is not unique. Example: d/dx (x2 + 2) = 2x ∫ 2xdx = x2 + C where C is the constant of integration. 5. (a) When a polynomial function is differentiated, the result is a polynomial function whose degree is one less than the degree of the given polynomial function. (b) When a polynomial function is integrated, then the result is a polynomial function whose degree is one more than the degree of the given polynomial function. Example: Let f(x) = x3 - 2x2 + 5x - 1 f ' (x) = 3x2 - 4x + 5 ∫f(x)dx = ∫(x3 - 2x2 + 5x - 1)dx = ∫x3 dx - 2 ∫x2 dx+ 5 ∫x dx - ∫ dx ∫f(x) dx = x4/4 - 2(x3/3) + 5(x2/2) - x + C = x4/4 - 2x3/3 + 5x2/2 - x + C ∫f(x) dx = x4/4 - 2x3/3 + 5x2/2 - x + C 6. The derivative of a function can be found at a point, whereas the integral of a function can be found over an interval. 7. (a) The geometrical meaning of derivative is the slope of the tangent to the corresponding curve at a point. (b) An indefinite integral can be represented geometrically. It is a family of curves placed parallel to each other and having parallel tangents at the points of intersection of the curves of the family with lines orthogonal to the axis representing the variable of integration. y = x2 + 2 y = x2 + 1 y = x2 y = x2 - 1 y = x2 - 2 On drawing the curves for these functions, we can observe that they are parallel to each other. 8. (a) Derivatives are useful in finding some physical quantity, like the velocity of a moving particle. (b) Integrals are useful in finding the displacement when the velocity of a particle at time t is given. Example: If the displacement of a moving particle is defined as s = 5t2 - 2t + 7 units, then find the velocity at t = 3 secs. Given, s = 5t2 - 2t + 7 units ⇒ ds/dt = 10t -2 v = ds/dt = 10t -2 [v]t=3sec = 10(3) - 2  = 30 -2  = 28 units/sec v  = 28 units/sec Example: If the velocity of a moving particle is given by the relation, v = 2 + 4t unts /sec, then find the displacement after 5 sec. Solution: Given that v = 2 + 4t units/sec v = 2 + 4t units/sec ∫v dt = ∫ (2 + 4t) dt Displacement, s = 2 ∫ dt + 4 ∫ t dt s = 2t + 4 x t2/2 + C s = 2t + 2t2 [S]t=5sec = 2(5) + 2(5)2  = 10 + 50  = 60 units Hence, the displacement after 5 seconds is 60 units. 9. The processes of both differentiation and integration involve limits. 10. The processes of differentiation and integration are inverses of each other. Example: d/dx(sin x) = cos x ∫ cos x dx = sin x + C

#### Summary

1. Differentiation and integration operate on functions. d/dx[p1f1(x) + p2f2(x)] = p1 d/dx f1(x) + p2 d/dx f2(x) ∫[p1f1(x) + p2f2(x)] dx = p1 ∫ f1(x) dx + p2 ∫ f2(x) dx 2. Differentiation and integration satisfy the property of linearity. a) Property of linearity in differentiation: d/dx[p1f1(x) + p2f2(x)] = p1 d/dx f1(x) + p2 d/dx f2(x) b) Property of linearity in integration: ∫[p1f1(x) + p2f2(x)] dx = p1 ∫ f1(x) dx + p2 ∫ f2(x) dx Let y = 10 log x + 4 sin x d/dx(y) = d/dx (10 log x + 4 sin x) = 10 d/dx (log x) + 4 d/dx (sin x) ∫[10/x + 4cos x] dx = 10 ∫1/x dx + 4 ∫ cos x dx Integrable function: If a function f(x) has an integral, then f(x) is called an integrable function. 3. All functions may not be differentiable, and all functions may not be integrable. Example: Prove that the function f(x) = |x - 2|, x ∈ R is not differentiable at x = 2 Given function is f(x) = |x - 2|, x ∈ R $\text{}\mathrm{| x-2 |}=\left\{\begin{array}{ll}x-2& \hfill \phantom{\rule{10}{0ex}}\mathrm{when x}\ge 2\\ -\left(x-2\right)& \hfill \phantom{\rule{10}{0ex}}\mathrm{when x}<2\end{array}\right\$ and LHL =  $\underset{h\to 0⁻}{\text{lim}}\frac{\text{f(2+h) - f(2)}}{\text{h}}$   = (-(2+ h - 2) - 0)/h = -h/h = -1 RHL = =  $\underset{h\to 0⁻}{\text{lim}}\frac{\text{f(2+h) - f(2)}}{\text{h}}$   = ((2 + h + 2) - 0)/h = h/h = 1 LHL ≠ RHL Hence,=    does not exist. Therefore, f is not differentiable at 2. Ex: Let f(x) = sinx / x ∫f(x)dx = ∫sinx/x dx d/dx(?) = sin x /x Hence, f(x) = sin x /x is not integrable. 4. The derivative of a function, if it exists, is unique. The integral of a function, if it exists, is not unique. Example: d/dx (x2 + 2) = 2x ∫ 2xdx = x2 + C where C is the constant of integration. 5. (a) When a polynomial function is differentiated, the result is a polynomial function whose degree is one less than the degree of the given polynomial function. (b) When a polynomial function is integrated, then the result is a polynomial function whose degree is one more than the degree of the given polynomial function. Example: Let f(x) = x3 - 2x2 + 5x - 1 f ' (x) = 3x2 - 4x + 5 ∫f(x)dx = ∫(x3 - 2x2 + 5x - 1)dx = ∫x3 dx - 2 ∫x2 dx+ 5 ∫x dx - ∫ dx ∫f(x) dx = x4/4 - 2(x3/3) + 5(x2/2) - x + C = x4/4 - 2x3/3 + 5x2/2 - x + C ∫f(x) dx = x4/4 - 2x3/3 + 5x2/2 - x + C 6. The derivative of a function can be found at a point, whereas the integral of a function can be found over an interval. 7. (a) The geometrical meaning of derivative is the slope of the tangent to the corresponding curve at a point. (b) An indefinite integral can be represented geometrically. It is a family of curves placed parallel to each other and having parallel tangents at the points of intersection of the curves of the family with lines orthogonal to the axis representing the variable of integration. y = x2 + 2 y = x2 + 1 y = x2 y = x2 - 1 y = x2 - 2 On drawing the curves for these functions, we can observe that they are parallel to each other. 8. (a) Derivatives are useful in finding some physical quantity, like the velocity of a moving particle. (b) Integrals are useful in finding the displacement when the velocity of a particle at time t is given. Example: If the displacement of a moving particle is defined as s = 5t2 - 2t + 7 units, then find the velocity at t = 3 secs. Given, s = 5t2 - 2t + 7 units ⇒ ds/dt = 10t -2 v = ds/dt = 10t -2 [v]t=3sec = 10(3) - 2  = 30 -2  = 28 units/sec v  = 28 units/sec Example: If the velocity of a moving particle is given by the relation, v = 2 + 4t unts /sec, then find the displacement after 5 sec. Solution: Given that v = 2 + 4t units/sec v = 2 + 4t units/sec ∫v dt = ∫ (2 + 4t) dt Displacement, s = 2 ∫ dt + 4 ∫ t dt s = 2t + 4 x t2/2 + C s = 2t + 2t2 [S]t=5sec = 2(5) + 2(5)2  = 10 + 50  = 60 units Hence, the displacement after 5 seconds is 60 units. 9. The processes of both differentiation and integration involve limits. 10. The processes of differentiation and integration are inverses of each other. Example: d/dx(sin x) = cos x ∫ cos x dx = sin x + C

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