Integration and Area Foundation Definite Integrals Todays Goals

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

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

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 Summing these strips gives an underestimate of the area that decreases as the number of strips increases 30 Strips dx=1/30 50 Strips dx=1/50 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 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 ∆x With increasing strips, ∆x ∆y tend to Zero and we can write and y Δx

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

As n increases the approximation of the area gets better Underestimate y 0 y 1 y 2 y 3 Overestimate ∆x Δx a) The area of shaded rectangle 7 is y 6. Δx ; (8 Strips here) The approximation of the total required Area is b) The area of shaded rectangle 7 is (y 6+dy 6). Δx is < Area < and of the total required area

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

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

In our specific example we have y y=x 2 = 0 = Foundation- Definite Integrals 1 x

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) 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

Example Area = 4½ square units Foundation- Definite Integrals

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

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