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Integrals - Evaluation of simple integrals and problems based on them

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: Integration is the reverse of differentiation. If the derivative of F(x)F(x) is f(x)f(x), then the integral of f(x)f(x) is F(x)+CF(x) + C. Visually, the constant of integration CC represents a family of curves that are vertical translations of each other, meaning they all share the same slope at any given xx-coordinate.

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Geometric Interpretation of Definite Integrals: A definite integral ∫abf(x)dx\int_{a}^{b} f(x) dx represents the signed area between the curve y=f(x)y = f(x) and the xx-axis from x=ax = a to x=bx = b. Geometrically, areas above the xx-axis are considered positive, while areas below are negative.

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Linearity Properties: The integral of a sum of functions is the sum of their 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, constants can be pulled out of the integral: ∫k⋅f(x)dx=k∫f(x)dx\int k \cdot f(x) dx = k \int f(x) dx. This allows complex polynomials to be integrated term-by-term.

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Integration by Substitution (Change of Variable): This method is used when an integrand contains a function and its derivative. By setting u=g(x)u = g(x), then du=g′(x)dxdu = g'(x) dx, we simplify the integral into a standard form. This is conceptually similar to the 'chain rule' in reverse.

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Fundamental Theorem of Calculus: This theorem provides a link between the antiderivative and the definite integral. It states that if F(x)F(x) is the antiderivative of f(x)f(x), then ∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x) dx = F(b) - F(a). This allows us to calculate areas without using Riemann sums.

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Standard Trigonometric Integrals: Basic trigonometric integrals are derived directly from differentiation rules. For example, since the derivative of tan⁡x\tan x is sec⁡2x\sec^2 x, the integral of sec⁡2x\sec^2 x is tan⁡x+C\tan x + C. Understanding the periodic nature of these functions helps in visualizing the areas they bound.

📐Formulae

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

∫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

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

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

💡Examples

Problem 1:

Evaluate the indefinite integral: ∫(3x2−4x+5)dx\int (3x^2 - 4x + 5) dx

Solution:

Step 1: Apply the linearity property to separate the terms: ∫3x2dx−∫4xdx+∫5dx\int 3x^2 dx - \int 4x dx + \int 5 dx. Step 2: Apply the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} to each term: ∫3x2dx=3⋅x33=x3\int 3x^2 dx = 3 \cdot \frac{x^3}{3} = x^3 ∫4x1dx=4⋅x22=2x2\int 4x^1 dx = 4 \cdot \frac{x^2}{2} = 2x^2 ∫5dx=5x\int 5 dx = 5x Step 3: Combine the results and add the constant of integration CC: x3−2x2+5x+Cx^3 - 2x^2 + 5x + C.

Explanation:

The solution uses the basic power rule and the linearity property of integrals to solve a polynomial integrand term-by-term.

Problem 2:

Evaluate: ∫2xcos⁡(x2)dx\int 2x \cos(x^2) dx

Solution:

Step 1: Use the substitution method. Let u=x2u = x^2. Step 2: Differentiate uu with respect to xx: dudx=2x\frac{du}{dx} = 2x, which gives du=2xdxdu = 2x dx. Step 3: Substitute uu and dudu into the original integral: ∫cos⁡(u)du\int \cos(u) du. Step 4: Integrate with respect to uu: sin⁡(u)+C\sin(u) + C. Step 5: Substitute the original value of uu back into the expression: sin⁡(x2)+C\sin(x^2) + C.

Explanation:

This problem is solved using the method of substitution because the derivative of the inner function x2x^2 (which is 2x2x) is present as a factor in the integrand.