Integration and Area Foundation- Definite Integrals

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

Remember from differentiation Foundation- Definite Integrals Remember from differentiation

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

Consider y=x 2 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 30 Strips dx=1/30 50 Strips dx=1/50 Summing these strips gives an underestimate of the area that decreases as the number of strips increases Foundation- Definite Integrals

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

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

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

a)The area of shaded rectangle 7 is y 6.Δx ; (8 Strips here) The approximation of the total required Area is ∆x b)The area of shaded rectangle 7 is (y 6 +dy 6 ).Δx and of the total required area is < Area < As n increases the approximation of the area gets better Underestimate Overestimate y0y0 y1y1 y2y2 y3y3 Δx

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

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

In our specific example we have = = x y=x y Foundation- Definite Integrals

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

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

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

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

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

Foundation- Definite Integrals Area = 4½ square units Example

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

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

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=y 2 +2