Area of a section bounded by a curve and two tangents

Question

I'm trying to solve the following problem:

Calculate the area of a section bounded by a curve $y=x^2+4x+9$ and two tangents in points: $x_1=-3$ and $x_2=0$.

I calculated the equations of the two tangents and I got that $y_1=-2x+0$ and $y_2=4x+9$.

Now I was able to draw the graph. But the question is, how to I calculate the area? I can take the integral

$\int_{-3}^{0} x^2+4x+9$, however, that would be the area of everything under the curve, not just the area of the section bounded by the two tangents.

How do I calculate the area of just the section?

Thanks

Answer 1

Those two tangent lines meet at $\left(-\frac32,3\right)$. So, in order to determine the area bounded by the graph and the two tangent lines. The tangent line corresponding to $x_1$ is $y=-2x$, whereas the tangent line corresponding to $x_2$ is $y=4x+9$.

So, all you have to do is to compute$$\int_{-3}^{-3/2}\overbrace{x^2+4x+9-(-2x)}^{\phantom{(x+3)^2}=(x+3)^2}\,\mathrm dx+\int_{-3/2}^0\overbrace{x^2+4x+9-(4x+9)}^{\phantom{x^2}=x^2}\,\mathrm dx.$$Each integral is equal to $\frac98$, and therefore your area is equal to $\frac94$.

Answer 2

Now, calculate $$2\int\limits_{-\frac{3}{2}}^{0}(x^2+4x+9-(4x+9))dx.$$

Answer 3

For calculation of area shift your x axis at y=3 Hence area will be area enclosed between parabola and , x=-3,x=0 and y=3 . From this substract area between tangents and y=3. Hence Required area is $$\int_{-3}^{0} (x^2+4x+6)dx -\frac{27}{4}$$ $$=33-\frac{27}{4}=\frac{105}{4}$$

Answer 4

There are many ways to find the area $[AED]$ in question, for example, we can use a bit of geometry:

\begin{align} [AED]&=[ABCD]-[ABC]-[DEC] . \end{align}

The equations of the tangent lines are \begin{align} f_1(x)&=f'(x_1)(x-x_1)+f(x_1) ,\\ f_2(x)&=f'(x_2)(x-x_2)+f(x_2) . \end{align}

So, given that

\begin{align} f'(x)&=2\,x+4 , \end{align}

we have

\begin{align} f_1(x)&=-2 x ,\\ f_2(x)&=4\,x + 9 , \end{align}

and we can find the coordinates of the corresponding points \begin{align} A&=(-3,6) ,\quad B=(-3,0) ,\quad C=(0,0) ,\quad D=(0,9) ,\quad E=(-\tfrac32,3) . \end{align}

Then corresponding areas are

\begin{align} [AED]&= \int_{x_1}^{x2} f(x)\,dx = \int_{-3}^{0} x^2+4\,x+9 \,dx =\left. \tfrac13\,x^3+2\,x^2+9\,x\right|_{x=-3}^{x=0} =18 ,\\ [ABC]&=\tfrac12\,|BC|\cdot|AC| =\tfrac12\cdot 3\cdot 6 =9 ,\\ [DEC]&= \tfrac12\,|CD|\cdot|E_x|= \tfrac12\,9\cdot\tfrac32= \tfrac{27}4 ,\\ [AED]&=18-9-\tfrac{27}4 =\tfrac94 . \end{align}