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Calculus - Indefinite and Definite Integration

Grade 12IB

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

🔑Concepts

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Indefinite Integration as Antidifferentiation: The indefinite integral ∫f(x)dx=F(x)+C\int f(x) dx = F(x) + C represents the collection of all functions whose derivative is f(x)f(x). Visually, the constant of integration CC creates a family of curves that are identical in shape but vertically translated up or down the y-axis.

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The Definite Integral and Area: The definite integral ∫abf(x)dx\int_{a}^{b} f(x) dx calculates the net signed area between the curve y=f(x)y=f(x) and the x-axis from x=ax=a to x=bx=b. Geometrically, regions where the function lies above the x-axis contribute positive values, while regions below the x-axis contribute negative values to the total integral.

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Fundamental Theorem of Calculus: This theorem links differentiation and integration, stating 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). Visually, this means the accumulated area under a rate-of-change graph is equal to the total change in the original function's value.

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Integration by Substitution (u-substitution): This technique is the reverse of the Chain Rule, used when an integral contains a function and its derivative, such as ∫f(g(x))g′(x)dx\int f(g(x))g'(x) dx. Visually, this can be thought of as a transformation of the x-axis to a new u-axis to simplify the geometry of the area being calculated.

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Area Between Two Curves: To find the area bounded by two functions f(x)f(x) and g(x)g(x) from x=ax=a to x=bx=b, we calculate ∫ab∣f(x)−g(x)∣dx\int_{a}^{b} |f(x) - g(x)| dx. Visually, this represents the 'vertical gap' between the top curve and the bottom curve summed across the interval.

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Volume of Revolution: When a region bounded by y=f(x)y=f(x) and the x-axis is rotated 360∘360^\circ around the x-axis, it forms a 3D solid. The volume is calculated using V=π∫ab[f(x)]2dxV = \pi \int_{a}^{b} [f(x)]^2 dx. Visually, this is modeled by summing an infinite number of thin cylindrical discs with radius R=f(x)R = f(x) and thickness dxdx.

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Properties of Definite Integrals: Integration is linear, meaning ∫(af(x)+bg(x))dx=a∫f(x)dx+b∫g(x)dx\int (af(x) + bg(x)) dx = a\int f(x) dx + b\int g(x) dx. Furthermore, if the limits are reversed, the sign of the integral changes: ∫abf(x)dx=−∫baf(x)dx\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx. Visually, switching limits is like 'reading' the area from right to left.

📐Formulae

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

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

∫exdx=ex+C\int e^x dx = e^x + C

∫axdx=axln⁡a+C\int a^x dx = \frac{a^x}{\ln a} + C

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

∫cos⁡(x)dx=sin⁡(x)+C\int \cos(x) dx = \sin(x) + C

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

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

V=π∫aby2dx(Volume around x-axis)V = \pi \int_{a}^{b} y^2 dx \quad \text{(Volume around x-axis)}

💡Examples

Problem 1:

Evaluate the indefinite integral ∫(3x2−4sin⁡(x)+e2x)dx\int (3x^2 - 4\sin(x) + e^{2x}) dx.

Solution:

  1. Integrate each term separately using the power rule, trigonometric rules, and exponential rules: ∫3x2dx=3⋅x33=x3\int 3x^2 dx = 3 \cdot \frac{x^3}{3} = x^3 ∫−4sin⁡(x)dx=−4(−cos⁡(x))=4cos⁡(x)\int -4\sin(x) dx = -4(-\cos(x)) = 4\cos(x) ∫e2xdx=12e2x\int e^{2x} dx = \frac{1}{2}e^{2x} (using u-substitution where u=2xu=2x)
  2. Combine the results and add the constant of integration CC: x3+4cos⁡(x)+12e2x+Cx^3 + 4\cos(x) + \frac{1}{2}e^{2x} + C

Explanation:

This problem demonstrates the linearity of integration (splitting the integral into parts) and the application of basic integration rules for polynomials, trigonometry, and exponentials.

Problem 2:

Find the area of the region enclosed by the curve y=x2y = x^2 and the line y=x+2y = x + 2.

Solution:

  1. Find the intersection points by setting the equations equal: x2=x+2  ⟹  x2−x−2=0  ⟹  (x−2)(x+1)=0x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0. The points are x=−1x = -1 and x=2x = 2.
  2. Identify the upper function. For x∈[−1,2]x \in [-1, 2], the line y=x+2y = x + 2 is above the parabola y=x2y = x^2.
  3. Set up the integral for the area: A=∫−12[(x+2)−x2]dxA = \int_{-1}^{2} [(x + 2) - x^2] dx.
  4. Integrate: [x22+2x−x33]−12[\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^{2}.
  5. Substitute the limits: F(2)=(42+4−83)=6−83=103F(2) = (\frac{4}{2} + 4 - \frac{8}{3}) = 6 - \frac{8}{3} = \frac{10}{3} F(−1)=(12−2−−13)=12−2+13=3−12+26=−76F(-1) = (\frac{1}{2} - 2 - \frac{-1}{3}) = \frac{1}{2} - 2 + \frac{1}{3} = \frac{3-12+2}{6} = -\frac{7}{6}
  6. Calculate the final area: 103−(−76)=206+76=276=4.5\frac{10}{3} - (-\frac{7}{6}) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = 4.5.

Explanation:

To find the area between curves, we first determine the boundaries via intersection, identify which curve is the 'ceiling' and which is the 'floor', and then integrate their difference.