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
Best Answer
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 $\frac94$.