Answer to Question #148272 in Calculus for rajitha

Question

Evaluate the Riemann sum for f(x)=6-2x find riemann sum under the area of curve with n=4 and equal subintervals [1,4] b)determine the actual under area under curve and explain why your answer in b underestimate

Answer 1

We will approximate.

$\intop_1^4 (6-2x)\delta x$

Note that $f(x)=6-2x$ and $a=1$ and $b=4$

$n=4$, so $\Delta x =\frac{b-a}{n}=\frac{4-1}{4}=\frac{3}{4}$ or $0.75$

All endpoints: start with $a$ and add $Δx$ successively: $1,\ 1.75,\ 2.5,\ 3.25,\ 4$

The sub-intervals are;

$$(1,1.75),\ (1.75, 2.5),\ (2.5, 3.25),\ (3.25, 4)$$

we will use the left endpoint of each sub-interval.

The left endpoints are;

$$1,\ 1.75,\ 2.5,\ 3.25$$

Now the Riemann sum is the area of the 4 triangles. We find the area of the rectangle by

$$Height \times Base = f(sample\ point) \times \Delta x$$

Here we are using left endpoints of sub-intervals for sample points and $\Delta x=0.70$, so

$$R=(f(1)+f(1.75)+f(2.5)+f(3.25)+f(4)) \cdot 0.75$$

$$R=(4+2.5+1+(-0.5)+(-2))\times 0.75=3.75 units ^2$$

b)determine the actual under area under curve and explain why your answer in b underestimate

The actual area under the curve is

$\intop _1^4 (6-2x)\delta x \implies [6x-x^2]_1^4\\ =[6(4)-(4)^2]-[6-1] \implies 8-5\\ \therefore Area= 3\ units^2$

The actual area is underestimated because I used the Left Riemann Sum and the function f(x) was decreasing on an interval