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Calculus - Integration

Grade 12IB_AI

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

πŸ”‘Concepts

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Integration is the inverse process of differentiation, often called finding the anti-derivative. If dydx=f(x)\frac{dy}{dx} = f(x), then y=∫f(x)dxy = \int f(x) dx.

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Indefinite Integration: When we find the general anti-derivative, we must add a constant of integration CC because the derivative of a constant is zero.

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The Power Rule: To integrate xnx^n, increase the power by 11 and divide by the new power. This is valid for all nβ‰ βˆ’1n \neq -1.

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Definite Integration: Used to find the exact numerical value between two limits aa and bb. It is written as ∫abf(x)dx\int_{a}^{b} f(x) dx and calculated as F(b)βˆ’F(a)F(b) - F(a).

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Area Under a Curve: The definite integral ∫abf(x)dx\int_{a}^{b} f(x) dx represents the area bounded by the curve y=f(x)y = f(x), the xx-axis, and the vertical lines x=ax=a and x=bx=b.

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Trapezoidal Rule: A numerical method used to approximate the area under a curve by dividing it into nn trapezoids of equal width h=bβˆ’anh = \frac{b-a}{n}.

πŸ“Formulae

∫xndx=xn+1n+1+C(nβ‰ βˆ’1)\int x^n dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)

∫kdx=kx+C\int k dx = kx + C

∫[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

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

Aβ‰ˆ12h[y0+yn+2(y1+y2+β‹―+ynβˆ’1)]A \approx \frac{1}{2}h [y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1})]

πŸ’‘Examples

Problem 1:

Find the indefinite integral: ∫(6x2βˆ’4x+3)dx\int (6x^2 - 4x + 3) dx.

Solution:

∫(6x2βˆ’4x+3)dx=6x33βˆ’4x22+3x+C=2x3βˆ’2x2+3x+C\int (6x^2 - 4x + 3) dx = \frac{6x^3}{3} - \frac{4x^2}{2} + 3x + C = 2x^3 - 2x^2 + 3x + C

Explanation:

Apply the power rule to each term individually. Increase the exponent by 1 and divide by the new exponent. Don't forget the constant CC.

Problem 2:

Evaluate the definite integral ∫13(3x2)dx\int_{1}^{3} (3x^2) dx.

Solution:

∫133x2dx=[x3]13=(3)3βˆ’(1)3=27βˆ’1=26\int_{1}^{3} 3x^2 dx = [x^3]_{1}^{3} = (3)^3 - (1)^3 = 27 - 1 = 26

Explanation:

First, find the anti-derivative of 3x23x^2, which is x3x^3. Then, substitute the upper limit (3) and the lower limit (1) and subtract the results.

Problem 3:

The velocity of a particle is given by v(t)=4t+1v(t) = 4t + 1 m/s. Find the displacement between t=0t = 0 and t=2t = 2 seconds.

Solution:

s=∫02(4t+1)dt=[2t2+t]02=(2(2)2+2)βˆ’(0)=8+2=10Β meterss = \int_{0}^{2} (4t + 1) dt = [2t^2 + t]_{0}^{2} = (2(2)^2 + 2) - (0) = 8 + 2 = 10 \text{ meters}

Explanation:

Displacement is the definite integral of velocity over the given time interval. Integrate 4t+14t + 1 to get 2t2+t2t^2 + t and evaluate between the limits.