Approximate area by an endpoint or trapezium rule

Which heights are used for a left-endpoint rectangle rule?

Try all eight subinterval counts for the function x squared plus one on the interval from zero to two. Watch the two outside sums and their average

What happens to the gap between right and left sums?

Use the rectangle rules

Partition the interval into equal subintervals and use one endpoint from each subinterval. Multiply the selected heights by the common width and add the rectangle areas

1. 1Find the common width Divide the interval length by the number of subintervals \(\displaystyle w=\frac{b-a}{n}\)
2. 2Choose the stated endpoint A left rule uses the first endpoint of each subinterval, while a right rule uses the second endpoint \(\displaystyle L=w\left(f(a)+f(a+w)+\cdots+f(b-w)\right)\)
3. 3Multiply each height by the width The area of each rectangle is its height multiplied by the common width \(\displaystyle A_i=w\times f(x_i)\)
4. 4Compare with the curve direction For an increasing function, left rectangles underestimate and right rectangles overestimate the area \(\displaystyle f\text{ increasing}\quad\Rightarrow\quad L<\int_a^b f(x)\,dx<R\)

Given

\(\displaystyle f(0)+f(1)+f(2)=3+5+7=\square\)

Evaluate and add the heights selected by the left rule

Given

\(\displaystyle 1\times15=\square\)

Multiply the height sum by the common width to find the area approximation

For three unit-width subintervals on [0,3], which expression uses the left endpoints?

Given

\(\displaystyle f(x)=x^2+1,\quad [0,2],\quad n=4,\quad h=\frac{2-0}{4}\)

Find the common width for four equal subintervals

Four neighbouring trapeziums use five endpoint heights. Which weighting is correct?

Add the trapezium areas

For f(x) = x² + 1 at x = 0, 1/2, 1, 3/2, 2, the heights are 1, 5/4, 2, 13/4, 5. Add neighbouring averages, or collect their shared endpoints into one weighted sum

1. 1Average each pair of endpoint heights The straight top joins the two endpoint heights. Its trapezium area is width times their average \(\displaystyle A_i=\frac{h}{2}(y_{i-1}+y_i)\)
2. 2Count shared endpoints twice An interior height belongs to the trapezium on each side. Each outer height belongs to only one \(\displaystyle T_n=\frac{h}{2}(y_0+2y_1+\cdots+2y_{n-1}+y_n)\)
3. 3Average the two rectangle sums Use the same partition and the same unrounded heights in both sums \(\displaystyle T_n=\frac{L_n+R_n}{2}\)

Given

\(\displaystyle S=1+2(5/4)+2(2)+2(13/4)+5\)

Calculate the weighted height sum before multiplying by h/2.

Given

\(\displaystyle h=1/2,\quad S=19,\quad T_4=(h/2)S\)

Apply the width and the factor 1/2.

Given

\(\displaystyle L_4=\frac12\left(1+\frac54+2+\frac{13}{4}\right)\)

Find the left sum on exactly the same partition

Given

\(\displaystyle R_4=\frac12\left(\frac54+2+\frac{13}{4}+5\right)\)

Find the right sum on exactly the same partition

For the same partition, why is a trapezium sum the average of the left and right sums?

Given

\(\displaystyle T_n=\frac{14}{3}+\frac{4}{3n^2},\quad n=8\)

Calculate the trapezium sum for eight equal subintervals using the supplied exact formula

A definite integral is the limiting sum

For this example, exact endpoint summation simplifies to the displayed formulas. As n grows, the terms 4/n and 4/(3n²) approach zero. All three sums therefore approach 14/3. This is an informal limit argument, not a requirement to use a finite number of strips as the exact answer

1. 1Refine the whole partition For a continuous function on a fixed closed interval, equal subintervals shrink as n increases. Rectangle and trapezium sums approach the definite integral; finite sums generally remain approximations \(\displaystyle h=\frac{b-a}{n}\longrightarrow0\)

Given

\(\displaystyle R_n-L_n=\frac8n,\quad n=8\)

Find the gap between the rectangle sums when n = 8.

Which statement about the trapezium sum and its limit is correct?

Increasing does not settle trapezium error

Both x² + 1 and 4x − x² increase on [0,2]. With two equal subintervals both give T₂ = 5. Their exact integrals, supplied for comparison, are 14/3 and 16/3 respectively. Thus the first trapezium estimate is above its integral and the second is below. Curvature matters

1. 1Separate direction from curvature Increasing behaviour brackets the integral between left and right sums. It does not alone say whether the trapezium sum is above or below the integral

A function is increasing and nonnegative. What can be concluded from that information alone on a shared partition?

Keep full values until the final entry

For calculator-based height evaluations, keep the unrounded stored values throughout the sum and apply the requested decimal precision once at the end. If a question explicitly supplies a rounded data table, use the table as instructed. Five listed endpoints create four strips; the width uses the strip count

1. 1Round only the final result Retain full stored heights and sums. Enter the final precision requested by a generated question; the exact lesson checks accept equivalent fractions or decimals

Apply the same weights to a reciprocal function

For f(x) = 6/(x+1) on [0,2], use two equal strips. The endpoint heights are exactly 6, 3 and 2, so no height rounding is needed. This positive function decreases; right rectangles give a lower estimate and left rectangles give an upper estimate

Given

\(\displaystyle f(x)=6/(x+1),\quad h=1,\quad(y_0,y_1,y_2)=(6,3,2)\)

Find the two-strip trapezium approximation for the reciprocal example