Integration and Area

1. Integration and Area Foundation- Definite Integrals

2. Today’s Goals • To understand why Integration can find the area under a Curve • To Introduce definite Integrals Recap We have already Introduced Integration as the inverse of Differentiation We know that we add a constant for INDEFINITE Integrals and that

3. Remember from differentiation Remember from differentiation Foundation- Definite Integrals

4. What is integration Integration is summing things up It means to piece things together , to add things up !! Foundation- Definite Integrals

5. Consider y=x2 If we wanted to find the area under this Curve between x=0 and x=1 we could use strips like this : 8 Strips dx=1/8 10 Strips dx=1/10 Summing these strips gives an underestimate of the area that decreases as the number of strips increases 50 Strips dx=1/50 30 Strips dx=1/30 Foundation- Definite Integrals

6. Summing these strips gives an overestimate of the area that decreases as the number of strips increases Foundation- Definite Integrals

7. Finding the Area If write the area of each of the individual rectangle as ∆A Then the area of each individual rectangle is between ∆y δy And So y y ∆x Δx With increasing strips, ∆x ∆y tend to Zero and we can write and

8. The area between a and b under the curve is Area ~ As the number of rectangles increases the approximation to the area improves Limit The Area under the curve from x=b to x=a , This is the Definite Integral and is written as Area = This Limit is written as The following slides explore this Foundation- Definite Integrals

9. As n increases the approximation of the area gets better Overestimate Underestimate y6+dy6 ∆y y6 8 7 6 ∆x ∆x y2 y3 Δx y1 y0 • The area of shaded rectangle 7 is y6.Δx ; (8 Strips here) • The approximation of the total required Area is • The area of shaded rectangle 7 is (y6+dy6).Δx and of the total required area is < Area <

10. < Area < where ∆A is difference in areas Taking an infinite amount of rectangles let n →∞ and so ∆x, ∆y, ∆A → 0 This RHS is the definition of the 1st derivative Limit Limit ∆X→0 Foundation- Definite Integrals

11. And So Change in Area With respect to (w.r.t) Change in x And so this leaves RHS because Limit ∆X→0 This is the value of x at the far side of the curve Using our knowledge of taking antiderivatives This is the value of x at the near side curve Foundation- Definite Integrals

12. In our specific example we have y y=x2 = x 1 0 = Foundation- Definite Integrals

13. Splitting Areas for Integration Where a curve is below the x-axis the integral is negative Therefore if the curve crosses the x axis we need to split the integration into seperate parts where |x| (the ‘modulus of x’) means the positive value of x

14. Example Find the area enclosed by the x axis and the curve y = 0 when x = 0 x = 2 and x = -1 The curve is below the axis for 0 < x < 2 and above the axis for -1 < x < 0

15. Example (needs checking!!) Find the area enclosed by the x axis and

16. Area between 2 curves f(x) and g(x) provided f(x) and g(x) don’t cross between a and b Foundation- Definite Integrals

17. Example First find the points of intersection of curve y=x2 and line y=x+2 At the points of intersection These will be our limits of integration Foundation- Definite Integrals

18. Example Area = 4½ square units Foundation- Definite Integrals

19. Area between the curve and the y axis To find this we rearrange the integral as d x δy c

20. Area between the curve and the y axis Rearrange as x=y2+2 Area = 12 2/3 square units

21. Splitting Areas for Integration We need to split integrals like this into 2 parts Because the Integral under the X-axes gives a negative result Finding Integrals between the Y axes and f(x) we rearrange the Integral as follows Rearrange as x=y2+2