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Integrals - Fundamental Theorem of Calculus

Grade 12CBSE

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

🔑Concepts

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The Area Function: For a continuous function ff on the interval [a,b][a, b], the area function is defined as A(x)=∫axf(t)dtA(x) = \int_a^x f(t) dt. Visually, this function represents the varying area under the curve y=f(t)y = f(t) from the fixed point aa to the moving point xx along the x-axis. As xx increases, the shaded region between the curve and the x-axis expands or contracts.

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First Fundamental Theorem of Calculus (FTC 1): This theorem establishes the link between differentiation and integration. It states that if ff is continuous on [a,b][a, b], then the area function A(x)A(x) is continuous on [a,b][a, b], differentiable on (a,b)(a, b), and A′(x)=f(x)A'(x) = f(x). This means the rate of change of the area under the curve at any point xx is equal to the height of the function at that point.

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Second Fundamental Theorem of Calculus (FTC 2): This theorem provides a practical method for evaluating definite integrals without using the limit of a sum. It states that if FF is any antiderivative of ff (i.e., F′(x)=f(x)F'(x) = f(x)), then ∫abf(x)dx=F(b)−F(a)\int_a^b f(x) dx = F(b) - F(a). Geometrically, this calculates the 'net' signed area between the graph of ff and the x-axis from x=ax = a to x=bx = b.

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Geometrical Interpretation: The definite integral ∫abf(x)dx\int_a^b f(x) dx represents the algebraic sum of the areas bounded by the curve y=f(x)y = f(x), the x-axis, and the vertical lines x=ax = a and x=bx = b. Visually, regions above the x-axis contribute positive area, while regions below the x-axis contribute negative area to the total integral value.

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Additive Property of Intervals: If ff is continuous on an interval containing a,b,a, b, and cc, then ∫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. Visually, this corresponds to splitting the total area under a curve into two adjacent sub-regions separated by a vertical partition line at x=cx = c.

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Linearity of the Definite Integral: The integral of a sum is the sum of the integrals, and constants can be factored out: ∫ab[k⋅f(x)+m⋅g(x)]dx=k∫abf(x)dx+m∫abg(x)dx\int_a^b [k \cdot f(x) + m \cdot g(x)] dx = k \int_a^b f(x) dx + m \int_a^b g(x) dx. This allows for the calculation of complex areas by breaking them down into simpler geometric components.

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Evaluation by Substitution: When solving ∫abf(g(x))g′(x)dx\int_a^b f(g(x)) g'(x) dx, we substitute u=g(x)u = g(x). Crucially, the limits of integration must be updated to g(a)g(a) and g(b)g(b). This process can be visualized as a transformation or 'stretching/shrinking' of the x-axis onto a new u-axis while maintaining the integrity of the area calculation.

📐Formulae

∫abf(x)dx=[F(x)]ab=F(b)−F(a)\int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a) where F′(x)=f(x)F'(x) = f(x)

ddx∫axf(t)dt=f(x)\frac{d}{dx} \int_a^x f(t) dt = f(x) (First Fundamental Theorem)

∫abf(x)dx=−∫baf(x)dx\int_a^b f(x) dx = -\int_b^a f(x) dx (Interchange of limits)

∫aaf(x)dx=0\int_a^a f(x) dx = 0

∫abk⋅f(x)dx=k∫abf(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

💡Examples

Problem 1:

Evaluate the definite integral: ∫12(3x2+2x+5)dx\int_1^2 (3x^2 + 2x + 5) dx

Solution:

Step 1: Find the antiderivative F(x)F(x) of the integrand f(x)=3x2+2x+5f(x) = 3x^2 + 2x + 5. Using the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1}: F(x)=3(x33)+2(x22)+5x=x3+x2+5xF(x) = 3\left(\frac{x^3}{3}\right) + 2\left(\frac{x^2}{2}\right) + 5x = x^3 + x^2 + 5x. Step 2: Apply the Second Fundamental Theorem of Calculus: ∫12f(x)dx=F(2)−F(1)\int_1^2 f(x) dx = F(2) - F(1). Step 3: Calculate F(2)F(2): F(2)=(2)3+(2)2+5(2)=8+4+10=22F(2) = (2)^3 + (2)^2 + 5(2) = 8 + 4 + 10 = 22. Step 4: Calculate F(1)F(1): F(1)=(1)3+(1)2+5(1)=1+1+5=7F(1) = (1)^3 + (1)^2 + 5(1) = 1 + 1 + 5 = 7. Step 5: Subtract the values: 22−7=1522 - 7 = 15.

Explanation:

This problem demonstrates the direct application of FTC 2. We first find the indefinite integral (antiderivative) and then evaluate the difference between the values at the upper and lower limits.

Problem 2:

Find the derivative of the function G(x)=∫0x1+t4dtG(x) = \int_0^{x} \sqrt{1 + t^4} dt with respect to xx.

Solution:

Step 1: Identify the components for the First Fundamental Theorem of Calculus. Here, f(t)=1+t4f(t) = \sqrt{1 + t^4} and the lower limit is a constant (00). Step 2: According to FTC 1, if G(x)=∫axf(t)dtG(x) = \int_a^x f(t) dt, then G′(x)=f(x)G'(x) = f(x). Step 3: Replace the variable tt in the integrand with the upper limit xx. Result: G′(x)=1+x4G'(x) = \sqrt{1 + x^4}.

Explanation:

This example utilizes the First Fundamental Theorem of Calculus (FTC 1). It shows that the derivative of an integral with respect to its upper limit is simply the value of the integrand at that upper limit, effectively 'undoing' the integration.