algebraic-curvesintegration
I'm really stuck on this. Say you have a curve $y = 3x – x^2$ that cuts the x-axis at points $O$ and $A$, and meets the line $y = -3x$ at the point $B$. How would you find the area of this shaded region?
I tried integrating between $O$ and $B$, and then I was going to take away the triangle (not shown) with hypotenuse $O$ to $B$. But this didn't work. The answer is 36, but I just can't get there.
We need the equation of the tangent line, whose slope is given by $m = f'(a)$, where $f'(a)$ is the derivative of the curve to which the line is tangent, evaluated at $x = a$.
Now, using the point $(a, b)$, the point of tangency, which happens to be a point on the line, to give us, along with slope, the equation of that line:
$$y - b = m(x-a) \iff y = mx + (b-ma)\iff y = m(x-a) + b$$
So we want the area under the line, evaluating the integral from $x = 0$ to the point where the line intersects the line $y = 0$, which is where $x = \dfrac{ma-b}{m} = a - \dfrac bm$
$$\int_0^{a - \large \frac bm} (m(x-a) +b)\,dx = \int_0^{a - \large\frac b{f'(a)}} \Big(f'(a)(x - a) + b\Big)\,dx$$
Hints:
$x_B$ can be found by solving $3x-x^2=0$. Now you can work out the area of the right triangle colored red in the figure.
Now the area of the shaded region can be represented as
$$A=\int_{x_A}^{x_B} 3x-x^2\,dx-\text{area of the triangle}$$
Best Answer
First evaluate the x value of $B$.
$$3x-x^2=-3x$$ $$6x-x^2=0$$ $$x(6-x)=0$$ So $B$ is at $x=6$ (and $A$ is at $x=0$)
Construct integral: $\int_0^63x-x^2\space-\space-3x\space dx$
$$\int_0^63x-x^2\space-\space-3x\space dx=\int_0^6 6x-x^2dx$$ $$=\left(3x^2-\frac{x^3}{3}\right)_0^6$$ $$=3\times6^2-\frac{6^3}{3}$$ $$=108-72$$ $$=36$$