Find the volume of the solid whose base is bounded by the graphs of y= x + 1 and y= x^2-1 with square cross sections taken perpendicular to the x-axis.
Thanks!
Thanks!
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The curves intersect at x = -1 and x = 2. Since the cross sections are squares, all we require is a formula for one side of the square as a function of x (squared of course, to get the area.)
V = ∫ A(x) dx from -1 to 2
Now, the base of the square is given by the distance between the two curves. Since the line is above the parabola, subtract the parabola from the line.
[x + 1] - [x^2 - 1]
- x^2 + 1 + x + 1
- x^2 + x + 2
But then, the area of a square will be that expression squared:
[- x^2 + x + 2][- x^2 + x + 2] = x^4 - 2 x^3 - 3 x^2 + 4 x + 4
A(x) = x^4 - 2 x^3 - 3 x^2 + 4 x + 4
Integrate that up now:
V = ∫ [x^4 - 2 x^3 - 3 x^2 + 4 x + 4] dx from -1 to 2
V = [(1/5)x^5 - (1/2)x^4 - x^3 + 2x^2 + 4x] from -1 to 2
Evaluate:
V = [(1/5)(2)^5 - (1/2)(2)^4 - (2)^3 + 2(2)^2 + 4(2)] - [(1/5)(-1)^5 - (1/2)(-1)^4 - (-1)^3 + 2(-1)^2 + 4(-1)]
V = 81/10
V = 8.1
Done!
Edit: Fixed a transcription error; answer remains unchanged.
V = ∫ A(x) dx from -1 to 2
Now, the base of the square is given by the distance between the two curves. Since the line is above the parabola, subtract the parabola from the line.
[x + 1] - [x^2 - 1]
- x^2 + 1 + x + 1
- x^2 + x + 2
But then, the area of a square will be that expression squared:
[- x^2 + x + 2][- x^2 + x + 2] = x^4 - 2 x^3 - 3 x^2 + 4 x + 4
A(x) = x^4 - 2 x^3 - 3 x^2 + 4 x + 4
Integrate that up now:
V = ∫ [x^4 - 2 x^3 - 3 x^2 + 4 x + 4] dx from -1 to 2
V = [(1/5)x^5 - (1/2)x^4 - x^3 + 2x^2 + 4x] from -1 to 2
Evaluate:
V = [(1/5)(2)^5 - (1/2)(2)^4 - (2)^3 + 2(2)^2 + 4(2)] - [(1/5)(-1)^5 - (1/2)(-1)^4 - (-1)^3 + 2(-1)^2 + 4(-1)]
V = 81/10
V = 8.1
Done!
Edit: Fixed a transcription error; answer remains unchanged.