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Calculus - Fundamental Theorem of Calculus

Grade 12ICSE

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

🔑Concepts

The Fundamental Theorem of Calculus (FTC) Part 1 connects differentiation and integration, showing they are inverse operations. If ff is continuous on [a,b][a, b], then the function g(x)=axf(t)dtg(x) = \int_{a}^{x} f(t) dt is continuous on [a,b][a, b] and differentiable on (a,b)(a, b), and g(x)=f(x)g'(x) = f(x). Visually, this means the rate of change of the 'area under the curve' from a fixed point aa to a moving point xx is equal to the height of the function at xx.

The Fundamental Theorem of Calculus Part 2 (Evaluation Theorem) provides a method for computing definite integrals. It states that abf(x)dx=F(b)F(a)\int_{a}^{b} f(x) dx = F(b) - F(a), where FF is any antiderivative of ff (i.e., F=fF' = f). Geometrically, this represents the net signed area bounded by the curve y=f(x)y = f(x), the x-axis, and the vertical lines x=ax = a and x=bx = b.

Area Interpretation: In the context of FTC, the definite integral calculates the cumulative area. If the graph of f(x)f(x) lies above the x-axis, the integral is the area. If the graph lies below the x-axis, the integral value is negative. The visual representation is a shaded region between the curve and the horizontal axis, bounded by the limits of integration.

Continuity Requirement: For the FTC to be valid, the integrand f(x)f(x) must be continuous on the entire closed interval [a,b][a, b]. If the function has a jump, an asymptote, or a hole within the boundaries, the theorem cannot be applied directly; the integral must instead be split at the point of discontinuity into separate intervals.

The Mean Value Theorem for Integrals: This concept states that for a continuous function ff on [a,b][a, b], there exists a point cc in (a,b)(a, b) such that f(c)=1baabf(x)dxf(c) = \frac{1}{b-a} \int_{a}^{b} f(x) dx. Visually, this means there is a rectangle with base (ba)(b-a) and height f(c)f(c) that has exactly the same area as the region under the curve.

Linearity and Additivity: Definite integrals follow the rules of linearity, where [f(x)+g(x)]dx\int [f(x) + g(x)] dx is the sum of individual integrals. Additionally, the 'interval addition' property allows us to split the area into two parts at any point cc: abf(x)dx=acf(x)dx+cbf(x)dx\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx. This is visually equivalent to dividing a single shaded region into two adjacent sub-regions.

Leibniz Rule for Differentiation under the Integral Sign: When the limits of an integral are functions of xx, say u(x)v(x)f(t)dt\int_{u(x)}^{v(x)} f(t) dt, the derivative is found by substituting the limits into the function and multiplying by their derivatives: f(v(x))v(x)f(u(x))u(x)f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x).

📐Formulae

ddx[axf(t)dt]=f(x)\frac{d}{dx} \left[ \int_{a}^{x} f(t) dt \right] = f(x) (FTC Part 1)

abf(x)dx=F(b)F(a)\int_{a}^{b} f(x) dx = F(b) - F(a) (FTC Part 2)

abkf(x)dx=kabf(x)dx\int_{a}^{b} k \cdot f(x) dx = k \int_{a}^{b} f(x) dx

ab[f(x)±g(x)]dx=abf(x)dx±abg(x)dx\int_{a}^{b} [f(x) \pm g(x)] dx = \int_{a}^{b} f(x) dx \pm \int_{a}^{b} g(x) dx

aaf(x)dx=0\int_{a}^{a} f(x) dx = 0

abf(x)dx=baf(x)dx\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx

ddx[g(x)h(x)f(t)dt]=f(h(x))h(x)f(g(x))g(x)\frac{d}{dx} \left[ \int_{g(x)}^{h(x)} f(t) dt \right] = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)

💡Examples

Problem 1:

Evaluate the definite integral: 13(3x2+4x+1)dx\int_{1}^{3} (3x^2 + 4x + 1) dx

Solution:

Step 1: Find the antiderivative F(x)F(x) of the integrand f(x)=3x2+4x+1f(x) = 3x^2 + 4x + 1. Using the power rule xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1}: F(x)=3(x33)+4(x22)+1(x)=x3+2x2+xF(x) = 3\left(\frac{x^3}{3}\right) + 4\left(\frac{x^2}{2}\right) + 1(x) = x^3 + 2x^2 + x Step 2: Apply the Fundamental Theorem of Calculus Part 2: 13(3x2+4x+1)dx=[x3+2x2+x]13\int_{1}^{3} (3x^2 + 4x + 1) dx = [x^3 + 2x^2 + x]_{1}^{3} Step 3: Substitute the upper limit (x=3x=3) and lower limit (x=1x=1): F(3)=(3)3+2(3)2+3=27+18+3=48F(3) = (3)^3 + 2(3)^2 + 3 = 27 + 18 + 3 = 48 F(1)=(1)3+2(1)2+1=1+2+1=4F(1) = (1)^3 + 2(1)^2 + 1 = 1 + 2 + 1 = 4 Step 4: Calculate the difference: 484=4448 - 4 = 44 Result: The value of the integral is 4444.

Explanation:

This problem demonstrates the direct application of FTC Part 2. We first find the general antiderivative and then find the difference between its values at the upper and lower boundaries.

Problem 2:

Find the derivative dydx\frac{dy}{dx} if y=0x2cos(t)dty = \int_{0}^{x^2} \cos(t) dt.

Solution:

Step 1: Identify the components for the Leibniz Rule/FTC Part 1. Here, the upper limit is v(x)=x2v(x) = x^2 and the lower limit is a constant a=0a = 0. The integrand is f(t)=cos(t)f(t) = \cos(t). Step 2: Use the formula ddxag(x)f(t)dt=f(g(x))g(x)\frac{d}{dx} \int_{a}^{g(x)} f(t) dt = f(g(x)) \cdot g'(x). Step 3: Substitute g(x)=x2g(x) = x^2 into the function f(t)f(t): f(x2)=cos(x2)f(x^2) = \cos(x^2) Step 4: Multiply by the derivative of the upper limit: ddx(x2)=2x\frac{d}{dx}(x^2) = 2x Step 5: Combine the results: dydx=cos(x2)2x=2xcos(x2)\frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x \cos(x^2) Result: dydx=2xcos(x2)\frac{dy}{dx} = 2x \cos(x^2).

Explanation:

This example shows how to differentiate an integral where the upper limit is a function of xx. We substitute the upper limit into the integrand and apply the chain rule by multiplying by the derivative of that limit.