A curve y=f(x) is passing through (2, 0) and slope of its tangent at any point (x, y) is (x+1)2+y−3x+1 , then the area bounded by the curve y=f'(x) and the axes is

Answer 1

This problem can begin by finding the function y=f(x), then finding y' and then finding the proper integral from 0 to 2, the given point. To find f(x) rearrange the problem to dy/dx+1=x²+5x. This can be solved by using an integrating factor = e^{integral dx}= e^x and multiplying all terms by e^x yielding d/dx(ye^x)=x(x+5)e^x. This can be solved by direct integration by parts to yield y=e^x+3x-3+ce^-x. C can be found by using the point (2,0) yield C=-3e²-e⁴. Now, find dy/dx= e^x+3-(3e²-e⁴)e^-x and integrate that from 0 to 2 finding area = 8-e²-e⁴