What is the are of the region ? Calculus?
What is the area of the region between the graphs of y = x^3 and y = -x - 1 from x = 0 to x = 2?
I'm not sure how to do these problem...
- IndikosLv 76 years agoFavorite Answer
We note that the graphs of y = x^3 lies above the x-axis
and the graph of y = -x - 1 lies below the x-axis.
Now the area of the graph of y = x^3 from x = 0 to x = 2
we determine as follows.
The integral of x^3 = x^/4.
So the required area above the x-axis = ( 2^4 - 0^4) /4 = 4.
Note that y= -x - 1 and the x-axis from x = 0 and x = 2
form a trapezium.
The sum of parallel sides = ( 1 + 3 ) = 4.
The height of the trapezium is 2.
So the area pf the trapz = sum of paralle sides * height /2
= 4 * 2 /2
So the required of region is area = 4 + 4
- JuventusLv 46 years ago
Area equal integral from 0 to 2 of ( x^3 - ( -x - 1 ) ) dx
= int ( x^3 + x + 1 ) dx from x=0 to 2
= [ (x^4/4) + (1/2 x^2) + x ] from x=0 to 2
Let f(x) = (x^4/4) + (1/2 x^2) + x
Thus, the integral = area = f(2) - f(0) = 8 - 0 = 8 squared unit.Source(s): B.Sc. in Applied Mathematics.