# Thread: Calculus Quiz Problems

1. ## Calculus Quiz Problems

Determine the area of the region enclosed by the graphs of f(x) = x^2 + x +5, g(x) = 3x +13, and the vertical line x = 0

I got 26.67 but he gave me zero points for it so idk.

Set up (but do not evaluate) an integral that will give the area of the surface generated when y= x^3/8 is revolved about the y-axis for 3 less than or equal to x, x less than or equal to 4.

2. ## Re: Calculus Quiz Problems

I would begin by plotting the region whose area we are to find:

We see the linear function is above the quadratic, and we know the lower limit of integration will be $\displaystyle x=0$. We need to determine where the upper limit is, and that occurs where the two functions are equal for $\displaystyle 0<x$.

$\displaystyle x^2+x+5=3x+13$

$\displaystyle x^2-2x-8=0$

$\displaystyle (x+2)(x-4)=0$

The positive root is:

$\displaystyle x=4$

And to the request area $\displaystyle A$ will be found from:

$\displaystyle A=\int_0^4 (3x+13)-(x^2+x+5)\,dx=\int_0^4 -x^2+2x+8\,dx=\left[-\frac{x^3}{3}+x^2+8x\right]_0^4=-\frac{64}{3}+16+32=\frac{80}{3}$

What you cited as the area appears to be a decimal approximation of this same value, rounded to two decimal places, so perhaps that's why you got no credit for it.