A curve passes through (2$$,$$0) and slope of the tangent at the point $$p(x$$,$$y)$$ is $$(x+1)2+y$$−$$3x+1$$, the area bounds by the curve and the x $$-axis$$ is

Question

A curve passes through (2$$,$$0) and slope of the tangent at the point $$p(x$$,$$y)$$ is $$(x+1)2+y$$−$$3x+1$$, the area bounds by the curve and the x $$-axis$$ is

Infinity Learn · Accepted Answer

$$dydx=x+1+y$$−$$3x+1$$, put $$X=x+1$$,$$Y=y$$−$$3$$⇒dYdX−$$YX=XI$$.$$F=e-$$∫$$1Xdx=e-log$$ $$X=1X$$ Solution of the D.E in $$YX=$$∫$$X1XdxYX=X+c$$⇒y−$$3x+1=x+1+c$$ $$----(1)$$ If $$x=2$$,$$y=0$$ ⇒  $$-33=3+c$$  ⇒$$c=-4sub$$ in (1)∴y−$$3=(x+1)2$$−$$4(x+1)=x2$$−$$2x$$−$$3$$⇒$$y=x2$$−$$2x$$ meets x $$-axis$$ ⇒$$y=0$$  $$x=0$$,$$2$$ Hence the required area $$=$$∫$$02$$ $$2x$$−$$x2dx=4$$−$$83=43$$

$A curve passes through (2,0) and slope of the tangent at the point p (x, y) i s \frac{(x + 1)^{2} + y - 3}{x + 1}, the area bounds by$
$the curve and the x -axis is$

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A curve passes through (2,0) and slope of the tangent at the point p(x,y) is (x+1)2+y−3x+1, the area bounds by the curve and the x -axis is

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$A curve passes through (2,0) and slope of the tangent at the point p (x, y) i s \frac{(x + 1)^{2} + y - 3}{x + 1}, the area bounds by$
$the curve and the x -axis is$