First let's find the intersection points of these curves
x2−3x=x ⟺ x2−4x=0x^2-3x=x\iff x^2-4x=0x2−3x=x⟺x2−4x=0
x=0;x=4x=0;\quad x=4x=0;x=4
The area between these curves can be found as a Riemann integral
S=∣∫04(x2−4x)dx∣=∣(x33−2x2)04∣=∣643−32∣=323S=\left|\int\limits_0^4(x^2-4x)dx\right|=\left|\left(\frac{x^3}{3}-2x^2\right)_0^4\right|=\left| \frac{64}{3}-32\right|=\frac{32}{3}S=∣∣0∫4(x2−4x)dx∣∣=∣∣(3x3−2x2)04∣∣=∣∣364−32∣∣=332
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment