Answer:
0.0032
Step-by-step explanation:
We need to compute
by the help of third-degree Taylor polynomial that is expanded around at x = 0.
Given :
< e < 3
Therefore, the Taylor's Error Bound formula is given by :
, where 



Therefore, |Error| ≤ 0.0032
Answer:
the last choice
Step-by-step explanation:
two functions are said to be inverse when they are symmetric about the line y=x
This is not picture so I don’t know what to do !
Answer:
Step-by-step explanation:
Approximate the integral
by dividing the region
with vertices (0,0),(4,0),(4,2) and (0,2) into eight equal squares.
Find the sum 
Since all are equal squares, so
for every 

Thus, 
Evaluating the iterate integral ![\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=\int\limits^4_0 {[xy+\frac{y^2}{2} ]}\limits^2_0 \, dx =\int\limits^4_0 {[2x+2]}dx\\\\=[x^2+2x]\limits^4_0=24.](https://tex.z-dn.net/?f=%5Cint%5Climits%5E4_0%20%5Cint%5Climits%5E2_0%20%7B%28x%2By%29%7D%20%5C%2C%20dydx%3D%5Cint%5Climits%5E4_0%20%7B%5Bxy%2B%5Cfrac%7By%5E2%7D%7B2%7D%20%5D%7D%5Climits%5E2_0%20%5C%2C%20dx%20%3D%5Cint%5Climits%5E4_0%20%7B%5B2x%2B2%5D%7Ddx%5C%5C%5C%5C%3D%5Bx%5E2%2B2x%5D%5Climits%5E4_0%3D24.)
Thus, 