The square root is throwing me of doing the integration. Can someone please show the steps of integration.
Answer is $\$26.19$
Ok then can someone show me how we got the integration here?
[Math] Consumer surplus for demand curve at the given sales level $x$
calculus
Related Solutions
The price $p$ when $200$ units are sold is, by the given formula, $1000-(0.1)(200)-(0.01)(200^2)$. Call this number $P$.
There would have been people willing to pay more than $P$. These people got a "deal," For the $\Delta x$ people willing to pay between $p(x)$ and $p(x+\Delta x)$, the amount "saved" is approximately $(p(x)-P)\Delta x$. "Add up" (integrate) from $x=0$ to $x=200$. The amount collectively saved (the consumer surplus) is $$\int_0^{200} (p(x)-P)\,dx.$$ For our particular problem, use the given function $p(x)$, and the value $P$ that we calculated. The integration will be straightforward.
Remark: Anyone unfortunate enough to have taught a calculus course in which this notion of marginal importance in economics is covered knows how to solve the problem. The notion of consumer surplus comes up not because it is significant, but because it is an "application" of the integral.
For context, I grabbed this picture from Wikipedia
The red area is the integral of $D(x) - 45$ from $0$ to $81$. Namely, $$ \int_0^{81} \left(\frac{405}{\sqrt{x}}-45\right)\,dx $$ To integrate, write $405/\sqrt{x}$ as $x^{-1/2}$ and use the formula $\int x^a = x^{a+1}/(a+1)$.
To check the answer, you can use Wolfram Alpha: it's $3645$.
Best Answer
The height of the curve at $x = 50$ is $\sqrt{16 - 0.14\times50}=\sqrt{9}=3$. Thus, the consumer surplus will be given by $$ \int_0^{50}(p(x)-50)dx = \int_0^{50}\left[\sqrt{16 - 0.14x}-3\right]dx $$ As Tyler indicated, find this integral using the $u$-substitution of $u = 16 - 0.14 x$.
So, we want the function whose derivative is $\sqrt{16 - 0.14x}$. If we naively just integrated the outside function, we would get $\frac{2}{3} (16 - 0.14x)^{3/2}$. Let's check what the derivative of this function would be. We find, using the chain rule, $$ \frac d{dx} \frac{2}{3}(16 - 0.14x)^{3/2} = \sqrt{16 - 0.14 x} \cdot(-0.14) $$ This is almost what we want, except that we have a $-0.14$ multiplying at the end. So, we're going to take what we got originally and divide by $-0.14$. We then have $$ \frac d{dx} \frac{1}{-0.14}\frac{2}{3}(16 - 0.14x)^{3/2} = \frac{1}{-0.14}\sqrt{16 - 0.14 x} \cdot(-0.14) = \sqrt{16 - 0.14 x} $$ And that's the derivative we wanted! So, we now know that $$ \int \sqrt{16 - 0.14 x} = -\frac{1}{0.14}\frac{2}{3}(16 - 0.14x)^{3/2} + C $$ Use this to find the integral you need.