<p>The mistake you are making appears to be treating $\theta_i$ as a <em>variable</em> while in reality it is a specific <em>value</em> of the derivative.</p>

<p>So 
$$\int_0^{q_i} V'(q)dq = \int_0^{q_i} \theta_i dq \;\;\; i=l,h$$</p>

<p>is wrong, since $V'(q)$ equals $\theta_l$ evaluated at $q_l$ only (and equals $\theta_h$ evaluated at $q_h$ only).</p>

CC BY-SA 3.0
2017-05-19T17:11:45.547Z
2017-05-19T17:35:50.893Z
2017-05-19T17:35:50.893Z
Alecos Papadopoulos
<p>I'm going to state the problem without much context because it is an issue of the math I'm doing, rather than any specific econ concept. </p>

<p>Condition. I'm trying to prove this.<br>
$$ V(q_l) - \theta_l*q_l  &gt; V(q_h) - \theta_h*q_h$$ </p>

<p>Given 
$$q_l&gt;q_h$$ 
$$\theta_l &lt; \theta_h  $$
$$V'() &gt; 0$$ $$ V'' &lt; 0 $$
$$V'(q_l) = \theta_l $$ 
$$V'(q_h) = \theta_h $$ </p>

<p>I'm trying to use definite integrals to prove this. But I think I'm getting something fundamentally wrong here. </p>

<p>$$ \int_0^{q_i} V'(q)dq = \int_0^{q_i} \theta_i  $$
$$ V(q_i) - V(0) = \theta_i*q_i$$ </p>

<p>This gives two equations, one for each $q_l, q_h$. Substracting these two equations from one another. </p>

<p>$$ V(q_l) - V(q_h) = \theta_l*q_l -\theta_h*q_h $$</p>

<p>I'm definitely doing something wrong here. Any help? </p>

<consumer-surplus>
Issues with an application of definite internal and marginal utilities
user795028
# Issues with an application of definite internal and marginal utilities

(Asked by: username_0 on 2017-05-19)

I'm going to state the problem without much context because it is an issue of the math I'm doing, rather than any specific econ concept.

Condition. I'm trying to prove this.  
$$ V(q\_l) - \theta\_l\*q\_l > V(q\_h) - \theta\_h\*q\_h$$

Given
$$q\_l>q\_h$$
$$\theta\_l < \theta\_h $$
$$V'() > 0$$ $$ V'' < 0 $$
$$V'(q\_l) = \theta\_l $$
$$V'(q\_h) = \theta\_h $$

I'm trying to use definite integrals to prove this. But I think I'm getting something fundamentally wrong here.

$$ \int\_0^{q\_i} V'(q)dq = \int\_0^{q\_i} \theta\_i $$
$$ V(q\_i) - V(0) = \theta\_i\*q\_i$$

This gives two equations, one for each $q\_l, q\_h$. Substracting these two equations from one another.

$$ V(q\_l) - V(q\_h) = \theta\_l\*q\_l -\theta\_h\*q\_h $$

I'm definitely doing something wrong here. Any help?

**Tags:** consumer-surplus

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## 1 answers

### Answer by username_1 on 2017-05-19. Score: 3 | Accepted answer

The mistake you are making appears to be treating $\theta\_i$ as a *variable* while in reality it is a specific *value* of the derivative.

So
$$\int\_0^{q\_i} V'(q)dq = \int\_0^{q\_i} \theta\_i dq \;\;\; i=l,h$$

is wrong, since $V'(q)$ equals $\theta\_l$ evaluated at $q\_l$ only (and equals $\theta\_h$ evaluated at $q\_h$ only).


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