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Integrals - Integration as inverse process of differentiation

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Integration as the Inverse Process of Differentiation: If the derivative of a function F(x)F(x) is f(x)f(x), then F(x)F(x) is called the anti-derivative or integral of f(x)f(x). Mathematically, if ddx[F(x)]=f(x)\frac{d}{dx}[F(x)] = f(x), then ∫f(x)dx=F(x)+C\int f(x) dx = F(x) + C.

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Constant of Integration (CC): Since the derivative of any constant is zero, different functions like x2+1x^2 + 1, x2−5x^2 - 5, and x2+100x^2 + 100 all have the same derivative 2x2x. To represent this entire family of functions, we add an arbitrary constant CC to the result of an indefinite integral.

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Geometric Interpretation: Visually, the indefinite integral ∫f(x)dx=F(x)+C\int f(x) dx = F(x) + C represents a family of curves. Each curve in the family is obtained by shifting one curve vertically along the y-axis. For a given value of x=ax = a, the tangents to all these curves are parallel, as they all have the same slope f(a)f(a).

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Integrand and Variable of Integration: In the notation ∫f(x)dx\int f(x) dx, the symbol ∫\int is the integral sign (an elongated 'S' for summation), f(x)f(x) is called the integrand, xx is the variable of integration, and dxdx indicates that the integration is performed with respect to xx.

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Linearity Property of Integrals: The integral of the sum or difference of two functions is the sum or difference of their individual integrals: ∫[f(x)±g(x)]dx=∫f(x)dx±∫g(x)dx\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx. Additionally, a constant factor can be moved outside the integral: ∫k⋅f(x)dx=k∫f(x)dx\int k \cdot f(x) dx = k \int f(x) dx.

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Comparison with Differentiation: While differentiation is a process used to find the rate of change or the slope of a tangent at a point, integration is used to recover the function when its slope (derivative) is known. Visually, differentiation 'breaks down' a function into its local slopes, while integration 'accumulates' these slopes to reconstruct the original function shape.

📐Formulae

∫xndx=xn+1n+1+C,n≠−1\int x^n dx = \frac{x^{n+1}}{n+1} + C, n \neq -1

∫dx=x+C\int dx = x + C

∫cos⁡xdx=sin⁡x+C\int \cos x dx = \sin x + C

∫sin⁡xdx=−cos⁡x+C\int \sin x dx = -\cos x + C

∫sec⁡2xdx=tan⁡x+C\int \sec^2 x dx = \tan x + C

∫csc⁡2xdx=−cot⁡x+C\int \csc^2 x dx = -\cot x + C

∫sec⁡xtan⁡xdx=sec⁡x+C\int \sec x \tan x dx = \sec x + C

∫csc⁡xcot⁡xdx=−csc⁡x+C\int \csc x \cot x dx = -\csc x + C

∫1xdx=log⁡∣x∣+C\int \frac{1}{x} dx = \log|x| + C

∫exdx=ex+C\int e^x dx = e^x + C

∫axdx=axlog⁡a+C\int a^x dx = \frac{a^x}{\log a} + C

∫11−x2dx=sin⁡−1x+C\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C

∫11+x2dx=tan⁡−1x+C\int \frac{1}{1+x^2} dx = \tan^{-1} x + C

💡Examples

Problem 1:

Find the anti-derivative (integral) of the function f(x)=3x2+4x3f(x) = 3x^2 + 4x^3 by the method of inspection.

Solution:

  1. We look for a function whose derivative is 3x23x^2. We know that ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2.
  2. We look for a function whose derivative is 4x34x^3. We know that ddx(x4)=4x3\frac{d}{dx}(x^4) = 4x^3.
  3. Combining these using the linearity of derivatives: ddx(x3+x4)=3x2+4x3\frac{d}{dx}(x^3 + x^4) = 3x^2 + 4x^3.
  4. Therefore, the anti-derivative is x3+x4+Cx^3 + x^4 + C.

Explanation:

The method of inspection involves identifying the 'parent' function through knowledge of standard differentiation formulas.

Problem 2:

Evaluate the integral: ∫(sin⁡x+cos⁡x)dx\int (\sin x + \cos x) dx.

Solution:

  1. Use the linearity property to split the integral: ∫sin⁡xdx+∫cos⁡xdx\int \sin x dx + \int \cos x dx.
  2. Apply the standard formula for ∫sin⁡xdx\int \sin x dx, which is −cos⁡x+C1-\cos x + C_1.
  3. Apply the standard formula for ∫cos⁡xdx\int \cos x dx, which is sin⁡x+C2\sin x + C_2.
  4. Combine the results and consolidate the constants: −cos⁡x+sin⁡x+C-\cos x + \sin x + C.

Explanation:

This example demonstrates how to use the sum rule of integration and standard trigonometric integral formulas.