Find the area under the graph & a curve approximate the area using rectangles. Practice upper and lower estimates, area under a curve calculus by rectangular approximation method.

From the earlier classes we have learned the formulas for finding the areas of some figures, which have only straight lines as sides. Some of these are given below.

(1)

(2)

(3)

(4)

It is not so easy to find the area of regions which have curved sides. We first start by approximating the area bounded by curved sides through rectangles.

Consider the curve \(\to\;y=x^3\)

Suppose we want to estimate the area under \(y=x^3\) from 0 to 1, base being \(x\) axis (shaded portion).

We can easily say that this area is less than that of square of length 1 so, shaded area < 1.

Since, area is a positive quantity, we say that \(0<A<1\).

A \(A>5\)

B \(0<A<1\)

C \(A>7\)

D \(A>2\)

We can get a closer estimate to the areas bounded by curved sides by breaking it into parts and then estimating each part by a rectangle.

Consider \(y=x^3\) in \([0,\,1]\)

We divide \([0,\,1]\) into two parts \(\to\;\left[0,\,\dfrac{1}{2}\right]\) and \(\left[\dfrac{1}{2},\,1\right]\)

Now, we observe that shaded area can be approximated by sum of two rectangles

\(\therefore\) Area of rectangles \(=\dfrac{1}{2}×\left(\dfrac{1}{2}\right)^3+\dfrac{1}{2}×1^3\)

\(=\dfrac{1}{2}×\dfrac{1}{8}+\dfrac{1}{2}×1\)

\(=\dfrac{1}{16}+\dfrac{1}{2}\)

\(=\dfrac{9}{16}=0.5625\)

\(\therefore\) Shaded Area \(=A<0.56\)

We have obtained a closer estimate to the area. We can get even better approximation by increasing the number of rectangles and ultimately an accurate area value by taking the limits.

- In this approximation, we have used the right end points as heights.

A \(A>2\)

B \(A<0.625\)

C \(A>-2\)

D \(A<0.1\)

We can get a closer estimate to the area bounded by curved sides by breaking it into parts and then estimating end part by a rectangle.

Consider \(y=x^6\) in \([0,\,1]\)

The area bounded is as shown.

Now this area can be approximated by dividing \([0,\,1]\) into four equal intervals \(\left[0,\,\dfrac{1}{4}\right],\;\left[\dfrac{1}{4},\,\dfrac{1}{2}\right],\;\left[\dfrac{1}{2},\,\dfrac{3}{4}\right]\) and \(\left[\dfrac{3}{4},\,1\right]\)

and calculating the area of rectangle formed.

\((A_1,\,A_2,\,A_3,\;\&\;\,A_4)\)

\(R_4=\) estimated area using four rectangles

\(=\dfrac{1}{4}×\underbrace{\left(\dfrac{1}{4}\right)^6}_{A_1}+\dfrac{1}{4}× \underbrace{\left(\dfrac{1}{2}\right)^6}_{A_2}+\dfrac{1}{4}×\underbrace{\left(\dfrac{3}{4}\right)^6}_{A_3}+\underbrace{\dfrac{1}{4}×1^6}_{A_4}\)

- This value will be larger than shaded area which was to be found.

\(\therefore\) We say that \(A<R_4\).

A \(A<0.3906\)

B \(A>5\)

C \(A=7\)

D \(A<-2\)

If we use the left end point height of the rectangles, we get a value which is a lower estimate of the area.

Consider \(y=x^3\) in [0,1].

We will use four rectangles.The area we calculate is shaded and is obviously less than the actual area.

\(L_4 = \dfrac{1}{4} × 0+\dfrac{1}{4}× \left(\dfrac{1}{4}\right)^3+\dfrac{1}{4}× \left(\dfrac{1}{2}\right)^3+\dfrac{1}{4}× \left(\dfrac{3}{4}\right)^3\)

(The left height of first rectangle is 0)

\(= \dfrac{1}{4} \left[\dfrac{1}{64}+\dfrac{1}{8}+\dfrac{27}{64}\right] \)

\(= \dfrac{1}{4}\left[\dfrac{1+8+27}{64}\right] \)

\(= \dfrac{9}{64} \)

\(=0 .1406\)

We say the area \(A>0 .1406\)

\(L_n\) will represent the area obtained by dividing the interval into n equal parts and using left end points for heights.

\(R_n\) = Shaded area

\(R_n>A\)

\(L_n\)= Shaded area

\(L_n<A\)

- As n increases, the values of \(R_n\) and \(L_n\) will approach the actual area.
- If the function is decreasing, we will observe that \(R_n<A\) and \(L_n >A\) .

A \(A>0.518\)

B \(A=7\)

C \(A=-2\)

D \(A=11\)

For an increasing function, we can get a closer estimate to the areas bounded by curved sides by breaking it into parts and then estimating each part by a rectangle.

Consider \(y=x^3\) in \([0,\,1]\)

We divide \([0,\,1]\) into two parts \(\to\;\left[0,\,\dfrac{1}{2}\right]\) and \(\left[\dfrac{1}{2},\,1\right]\). Now, we observe that shaded area can be approximated by sum of two rectangles.

\(\therefore\) Area of rectangles \(=\dfrac{1}{2}×\left(\dfrac{1}{2}\right)^3+\dfrac{1}{2}×1^3\)

\(=\dfrac{1}{2}×\dfrac{1}{8}+\dfrac{1}{2}×1\)

\(=\dfrac{1}{16}+\dfrac{1}{2}\)

\(=\dfrac{9}{16}=0.5625\)

\(\therefore\) Shaded Area \(=A<0.56\)

We have obtained a closer estimate to the area. We can get even better approximation by increasing the number of rectangles and ultimately an accurate area value by taking the limit.

- In this approximation, we have used the right end points as heights.

If we use the left end point height of the rectangles we get a value which is a lower estimate of the area.

Consider \(y=x^3\) in [0,1].

We will use four rectangles. The area we calculate is shaded and is obviously less than the actual area.

\(L_4 = \dfrac{1}{4} × 0+\dfrac{1}{4}× \left(\dfrac{1}{4}\right)^3+\dfrac{1}{4}× \left(\dfrac{1}{2}\right)^3+\dfrac{1}{4}× \left(\dfrac{3}{4}\right)^3\)

(The left height of first rectangle is 0)

\(= \dfrac{1}{4} \left[\dfrac{1}{64}+\dfrac{1}{8}+\dfrac{27}{64}\right] \)

\(= \dfrac{1}{4}\left[\dfrac{1+8+27}{64}\right] \)

\(= \dfrac{9}{64}\)

\( = .1406\)

We say the area \(A> .1406\)

\(L_n\) will represent the area obtained by dividing the interval into n equal parts and using left end points for heights.

\(R_n\)= Shaded Area

\(R_n>A\)

\(L_n\)= Shaded Area

\(L_n <A\)

- As n increases the values of \(R_n\) and \(L_n\) will approach the actual area.
- If the function is decreasing, we will observe that \(R_n<A\) and \(L_n >A\) .

A \(32<A<46\)

B \(A>49\)

C \(2<A<4\)

D \(-1<A<18\)

If \(f\) is a decreasing function , then \(R_n\) will give the lower estimate and \(L_n\) will give the upper estimate of the actual area.

Estimation using \(R_n=Shaded\;Area\)

Estimation using \(L_n=Shaded\; Area\)

\(R_n<A<L_n\)

A \(54<A<72\)

B \(A>96\)

C \(A<34\)

D \(2<A<3\)

Suppose instead of graphs we are given an expression of the function. We sketch that (using standard graphs and graphical transformation) and then approximate using \(R_n \,\,{\text{or}}\,\, L_n\) as the case may be

- If a function is not increasing or decreasing in an interval there is no fixed relation between \(R_n,\,L_n \,\,{\text {and}}\,\,A\).

A \(A\cong11.315\)

B \(A\cong52\)

C \(A\cong-\dfrac{1}{2}\)

D \(A\cong.01\)