The total area enclosed by the graphs of
y=10x^2-x^3+x
y=x^2+19x
Please Please help me solve this. I have done it amny times, but getting it wrong
To find the total area enclosed by the graphs of the given equations, you need to determine the points of intersection between the two curves. The enclosed area will be the definite integral of the difference between the two equations over the interval where they intersect.
To start, set the two equations equal to each other and solve for x:
10x^2 - x^3 + x = x^2 + 19x.
Rearrange the equation to obtain a cubic equation:
x^3 + 9x^2 + 18x = 0.
Factoring out an x:
x(x^2 + 9x + 18) = 0.
Setting each factor equal to zero:
x = 0,
x^2 + 9x + 18 = 0.
The quadratic equation x^2 + 9x + 18 = 0 does not factor, so you can solve it using the quadratic formula:
x = (-9 ± √(9^2 - 4(1)(18))) / (2(1)).
Simplifying the equation, you get:
x = (-9 ± √(81 - 72)) / 2
x = (-9 ± √9) / 2
x = (-9 ± 3) / 2.
This gives two possible values for x:
x1 = (-9 + 3) / 2 = -6/2 = -3,
x2 = (-9 - 3) / 2 = -12/2 = -6.
Now that you have the x-values, you can substitute them back into one of the original equations to find the corresponding y-values. Let's use the first equation:
For x = -3:
y = 10(-3)^2 - (-3)^3 + (-3)
y = 90 + 27 - 3 = 114.
For x = -6:
y = 10(-6)^2 - (-6)^3 + (-6)
y = 360 + 216 - 6 = 570.
Now you have the points of intersection: (-3, 114) and (-6, 570).
To calculate the area enclosed between the graphs, you need to integrate the difference between the two equations over the interval from -6 to -3:
Area = ∫[from -6 to -3] (10x^2 - x^3 + x) - (x^2 + 19x) dx.
Evaluating this integral will give you the total area enclosed by the graphs of the given equations. You can solve this either analytically or with the help of numerical techniques such as numerical integration or software like WolframAlpha or MATLAB.