Subject GRE math exam 9768 Q.21

Question

The question is given in the following picture:

I started to solve it first by differentiating $ y = \frac{1}{8} x^2 + \frac{1}{2}x + 1$, and I got $y^{'} = \frac{1}{4}x + \frac{1}{2}$, at point (0,1) it becomes $y^{'} = \frac{1}{2}$ and this is the slope of the line tangent to the given graph at (0,1).

So the equation of the line tangent to the given graph is $ y = \frac{1}{2}x + b $, now to determine b we use the given point (0,1), so $b = 1$.

Now I know that to solve the question I must solve the integral: $$\int_{0}^{1} (\frac{1}{2}x + 1)dx$$ But after finding its boundaries (by finding the x & y intercept) and solving it my answer was $\frac{5}{4}$ which was not one of the choices.

The correct answer is D. could anyone help me find my misconception or my mistake?

Thanks in advance.

Answer 1

The mistake is that $x$ varies between $-2$ and $0$, not between $0$ and $1$. The second interval is where $y$ is located, but in your calculation you're integrating with respect to $x$, hence you need to integrate from $-2$ to $0$.

Maybe a clever way to do it and to avoid integrals is to note that the area is a right-angled triangle with sides $2$ and $1$ and hence the area is $\frac{2 \cdot 1}{2} = 1$

Answer 2

The tangent line is indeed $y=1+x/2$, now not that the desired area is the triangle with vertices $(-2,0), (0,1), (0,0)$, hence $$ S_{\Delta} = \frac{2\times1}{2} = 1. $$