Splitting up the interval of integration into 
subintervals gives the partition
![\left[0,\dfrac1n\right],\left[\dfrac1n,\dfrac2n\right],\ldots,\left[\dfrac{n-1}n,1\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac1n%5Cright%5D%2C%5Cleft%5B%5Cdfrac1n%2C%5Cdfrac2n%5Cright%5D%2C%5Cldots%2C%5Cleft%5B%5Cdfrac%7Bn-1%7Dn%2C1%5Cright%5D)
Each subinterval has length 
. The right endpoints of each subinterval follow the sequence
with 
. Then the left-endpoint Riemann sum that approximates the definite integral is
and taking the limit as 
gives the area exactly. We have
F(x)=x^2 +8x
f(d-3) basically what we get from this is that the x value is d-3
to find f(d-3) simply plug in d-3 for every x
f(d-3)=(d-3)^2 + 8(d-3)
that would be one form of the answer, but we can also continue multiplying it out
f(d-3)= d^2-6d+9+8d-24
add like terms
f(d-3)=d^2+2d-15
have a nice day and i hope this helps :)
Answer:
Total area of figure = [3x² + 7x + 6] feet²
Step-by-step explanation:
By dividing both rectangle by vertical line;
Given:
Length of big rectangle = (x + 3) feet
Width of big rectangle = (x + 2) feet
Length of small rectangle = 2x feet
Width of small rectangle = (x + 1) feet
Find:
Total area of figure
Computation:
Area of rectangle = Length x width
Total area of figure = Area of big rectangle + Area of small rectangle
Total area of figure = [(x + 3)(x + 2)] + [(2x)(x + 1)]
Total area of figure = [x²+ 2x + 3x + 6] + [2x² + 2x]
Total area of figure = [3x² + 7x + 6] feet²
Answer:
there is none
Step-by-step explanation:
Your answer would be x= -42
