# 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...

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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

= 4.

So the required of region is area = 4 + 4

= 8.

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• 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.
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