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Calculus - Basic Integration and Area Under a Curve

Grade 12IGCSE

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

🔑Concepts

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Integration as the inverse process of differentiation (finding the anti-derivative).

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Indefinite Integration: Finding a general family of functions, characterized by the constant of integration (+ C).

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Definite Integration: Evaluating the integral between two specific limits to find a numerical value.

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Geometric Interpretation: The definite integral represents the signed area between the curve y = f(x) and the x-axis.

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Area below the x-axis: If the curve is below the x-axis, the integral value will be negative; the absolute value must be taken for area calculations.

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Fundamental Theorem of Calculus: Connecting the derivative and the integral.

📐Formulae

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

∫k dx=kx+C\int k \, dx = kx + C

∫(ax+b)ndx=(ax+b)n+1a(n+1)+C\int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n+1)} + C

∫abf(x)dx=[F(x)]ab=F(b)−F(a)\int_{a}^{b} f(x) dx = [F(x)]_{a}^{b} = F(b) - F(a)

Area = ∫aby dx\int_{a}^{b} y \, dx

💡Examples

Problem 1:

Find the indefinite integral: ∫(6x2−4x+3)dx\int (6x^2 - 4x + 3) dx.

Solution:

∫(6x2−4x+3)dx=6(x33)−4(x22)+3x+C=2x3−2x2+3x+C\int (6x^2 - 4x + 3) dx = 6(\frac{x^3}{3}) - 4(\frac{x^2}{2}) + 3x + C = 2x^3 - 2x^2 + 3x + C

Explanation:

Apply the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} to each term independently and don't forget the constant of integration C.

Problem 2:

Evaluate the definite integral: ∫13(4x+1)dx\int_{1}^{3} (4x + 1) dx.

Solution:

[4x22+x]13=[2x2+x]13=(2(3)2+3)−(2(1)2+1)=(18+3)−(2+1)=21−3=18[\frac{4x^2}{2} + x]_{1}^{3} = [2x^2 + x]_{1}^{3} = (2(3)^2 + 3) - (2(1)^2 + 1) = (18 + 3) - (2 + 1) = 21 - 3 = 18

Explanation:

First, find the anti-derivative, then substitute the upper limit (3) and subtract the result of substituting the lower limit (1).

Problem 3:

Calculate the area bounded by the curve y=x2y = x^2, the x-axis, and the lines x=0x = 0 and x=3x = 3.

Solution:

Area = ∫03x2dx=[x33]03=(333)−(033)=273−0=9\int_{0}^{3} x^2 dx = [\frac{x^3}{3}]_{0}^{3} = (\frac{3^3}{3}) - (\frac{0^3}{3}) = \frac{27}{3} - 0 = 9 square units.

Explanation:

The area is the definite integral of the function from the start point to the end point on the x-axis. Since x2x^2 is always positive in this interval, the integral gives the area directly.