Integral Calculus Solutions
Topics || Problems
Find the area bounded by \(y^2=4x\) and \(y+2x=12\)
Solution
Solution
I. Graph the curves.
II. Calculate the intersections between curves.
\(y=12-2x\)
\((12-2x)^2 = 4x\)
\(144-48x+4x^2 = 4x\)
\(144-52x+4x^2 = 0\)
\(36-13x+x^2 = 0\)
\(x_1 = 9\)
\(x_2=4\)
Solve the the corresponding values of \(y\).
At \(x = 9\)
\(y = 12-2(9) = -6\)
At \(x = 4\)
\(y = 12-2(4) = 4\)
III. Select the differential rectangle(use horizontal rectangle).
The differential area: \(dA=x dy\) but \(x = x_2-x_1\)
\(x_2=\frac{12-y}{2}\) : simplified equation of the line
\(x_1 = \frac{y^2}{4}\)
\(dA = [(\frac{12-y}{2})-(\frac{y^2}{4})]dy\)
IV. Integrate from the lower limit to the upper limit.
Lower limit:\(y=4\)
Upper limit:\(y=-6\)
\(A = \int_{-6}^4{(\frac{12-y}{2} -\frac{y^2}{4} \,dy )}\)
\(A =6y-\frac{y^2}{4}-\frac{y^3}{12}|_{-6}^4 \)
\(A = -(-36-9+18)+(24-4-5.333)\)
\(A = 41.67 \text{sq. units}\)