Area of a section bounded by a curve and two tangents 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
 A: 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 $\frac94$.
A: Now, calculate $$2\int\limits_{-\frac{3}{2}}^{0}(x^2+4x+9-(4x+9))dx.$$
A: 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 
$$\int_{-3}^{0} (x^2+4x+6)dx -\frac{27}{4}$$
$$=33-\frac{27}{4}=\frac{105}{4}$$
A: 
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} 
