Answer:
E
Step-by-step explanation:
Since the given expression is an elliptic integral, it's not easy to find F(x) explicitly. However, if we use a linear approximation, we can try to estimate F(3), though it won't be too close, since 3 is not very close to 1.
At the point (1,5) the slope of the tangent line is F'(1) = \(\sqrt{2}\)
So, using that line,
F(3) = F(1) + F'(1)(3-1) = 5+2\(\sqrt{2}\) = 7.828
Well, that didn't work out too well.
So, let's pull out our handy integral calculator, and we find that
\(\int\limits^0_1 {\sqrt{1+x^3} } \, dx = 1.111\)
But, we know that F(1) = 5. So we need to add C = 3.889
Now, integrating again,
\(\int\limits^0_3 \sqrt{1+x^3} \, dx = 7.341 + 3.889 = 11.229\)
Oops. I messed up the limits, but you get the idea.
