# Consumer surplus definition

Varian defines consumer surplus at price $$p = p_0$$ as $$\displaystyle CS(p_0) = \int_{p_0}^{p_\max} q(p) \ dp$$ where $$p_\max$$ exists and $$q(p_\max) = 0$$.

The other definition that I generally use is $$\displaystyle CS(p_0) = \int_{0}^{q_0} [p(q) - p_0] \ dq$$.

Are the two definitions same? I tried equating them and I couldn't prove that they are equal. Is the second definition incorrect?

Yes they are the same definitions. Graphically the different integrals calculate the same area just rotated 90 degrees. Geometric shapes have the same area regardless of rotation.

General Proof:

Suppose we have a region of fixed size called CS in I quadrant of Cartesian coordinate system enclosed by $$p(q)$$ and $$p= p_0$$. The area of the region would be given by:

$$\text{ CS} = \lim_{n \to \infty} \sum [p(q_i)- p_0] \Delta q, \text{with } \quad \Delta q = \frac{q_0-0}{n}$$

Taking the limit we get:

$$\text{CS} = \int_{0}^{q_0} [p(q) - p_0] \ dq$$

Now integrate the same fixed area CS bounded by $$p(q)$$ and $$p_0$$ by $$y$$ axis.

Since CS is bounded by $$p(q)$$ and $$p_0$$ on $$y$$ axis we are integrating over interval $$[p_{max},p_0]$$ over inverse function $$p(q)^{-1}$$ i.e. $$q=p(q)$$ (we do not include $$p_0$$ as it is constant function and thus the boundary on $$y$$ axis). Hence, on $$y$$ axes the fixed area of $$CS$$ is defined as:

$$\text{CS}= \lim_{n \to \infty} \sum q(p_i)\Delta p, \text{with} \quad \Delta p = \frac{p_{max}-p_0}{n}$$

taking the limit we get:

$$\text{CS}= \int_{p_0}^{p_{max}}q(p)dp$$

Since we are talking about $$CS$$ of the same size we have:

$$\int_{0}^{q_0} [p(q) - p_0] \ dq = \text{CS} = \int_{p_0}^{p_{max}}q(p)dp$$

And hence we proven that:

$$\int_{0}^{q_0} [p(q) - p_0] \ dq = \int_{p_0}^{p_{max}}q(p)dp$$

Example:

You can see that by trying various demand functions. Suppose we have demand given by $$Q=100-p$$, equilibrium price is given by $$p=10$$.

Varian:

$$CS(10)= \int_{10}^{100}(100-p)dp= 100(100) -\frac{1}{2}(100)^2-100(10) -\frac{1}{2}(10)^2=4050$$

Non-Varian:

we first have to solve for inverse demand which is given by: $$p=100-q$$ (recall that fixing \$p=10 \implies q^*=90).

$$CS(10)= \int_{0}^{90}[(100-q)-10]dp= 4050$$

This will hold for any arbitrary function since consumer surplus is graphically a geometrical object and the respective integrals are just different ways of calculating area of this object depending on how you rotate the plane with the object.

More generally what Varian approach is doing would be known as integrating along the y-axis.

• Can you please help me show that mathematically? I understand the rotation, but I don't see how to write that in a more rigorous manner. Oct 27, 2022 at 17:33
• @marblegoolies I updated the question with general proof for arbitrary p(q) and p_0, (of course assuming p(q) is continuous, inverse exists and is also continuous)
– 1muflon1
Oct 27, 2022 at 18:52
• Okay, I thought about 1. and I concede the point. Oct 27, 2022 at 22:03
• Would prefer your answer if the general proof came before the example, but to each her own. There you go, peer review (: Oct 27, 2022 at 22:06