Integration with limits
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This document introduces integration with limits. It shows how to integrate simple functions without limits and with limits. When integrating with limits, the function is integrated as normal and then the upper and lower limits are substituted and subtracted, with no need for the +c term. Several examples integrate functions from limits of 1 to various numbers to demonstrate this process. The key question asks to find the value of a if the integral of (2x + 6) from 0 to a equals 27.
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GCSEYr9-SolvingQuadratics.pptx
Here are the steps to solve this equation by factorising:
1) Factorise the left hand side: 3(x - 1)(x + 2)
2) Set each factor equal to 0:
x - 1 = 0
x + 2 = 0
3) Solve for x:
x = 1
x = -2
Therefore, the solutions are x = 1 or x = -2.
Single page-integral-table
1) This document provides formulas for integrals of common functions including polynomials, rational functions, radicals, logarithms, trigonometric functions, exponentials, and hyperbolic functions.
2) The integrals are organized into sections based on the type of function, with the formula provided along with conditions on parameters.
3) There are over 100 integrals presented in the table spanning multiple pages of formulas.
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Vedic Mathematics- Combined operation Sum and Product of Numbers
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Integration with limits
- 1. Block 2 Integration with Limits
- 2. What is to be learned? • What these limit things are and how they work
- 3. Introducing Limits ∫ 3x2 dx no limits 1 3 = x3 [ ] 1 3 = 33 13 = 26 +c +c – c Integrate Substitute Subtract –
- 4. ∫ 6x dx 1 4 = 3x2 [ ] 1 4 = 3(4)2 – 3(1)2 = 45
- 5. ∫ 8x3 dx 1 2 = 2x4 [ ] 1 2 = 2(2)4 – 2(1)4 = 30
- 6. Integrals With Limits Have wee numbers on them Integrate as normal Substitute wee numbers, then subtract No need for + c
- 7. ∫ 8x dx 1 5 = 4x2 [ ] 1 5 = 4(5)2 – 4(1)2 = 96 Ex 1.
- 8. ∫ (4x + 2) dx 1 3 = 2x2 + 2x[ ] 1 3 = 2(3)2 + 2(3) = 24 – 4 – ( 2(1)2 + 2(1) ) = 20 subtract all of this
- 9. ∫ (3x2 + 4) dx 1 3 = x3 + 4x[ ] 1 3 = (3)3 + 4(3) = 39 – 5 – ( (1)3 + 4(1) ) = 34 subtract all of this Ex 2.
- 10. ∫ (6x + 5) dx 1 2 = 3x2 + 5x[ ] 1 2 = 3(2)2 + 5(2) = 22 – 8 – ( 3(1)2 + 5(1) ) = 14 subtract all of this Key Question
- 11. ∫ (2x + 6) dx = 27 Find a (a > 0) 0 a = x2 + 6x[ ] o a = a2 + 6a – ( (0)2 + 6(0) ) Nasty = 27 Quadratic Equation a2 + 6a – 27 = 0 (a + 9)(a – 3) = 0 a = -9 or a = 3