The marginal revenue function for a firm is given by $\frac{5x^2+30x+51}{(x+3)^2}$. Show that the revenue function is given by $\frac{2x}{x+3}+5x$.

> [!TLDR] > Integrate $MR$ and fix the constant with $R(0)=0$; the result equals $\frac{2x}{x+3}+5x$. > [!FORMULA] > **Revenue from marginal revenue:** $R(x)=\int MR\,dx$, with the boundary condition $R(0)=0$ (no output ⇒ no revenue). 1. **Given:** $MR=\dfrac{5x^2+30x+51}{(x+3)^2}$. 2. **Rewrite the numerator about $(x+3)$:** $5x^2+30x+51=5(x^2+6x+9)+6=5(x+3)^2+6$. 3. So $MR=\dfrac{5(x+3)^2+6}{(x+3)^2}=5+\dfrac{6}{(x+3)^2}$. 4. **Integrate:** $R=\int\Big(5+\dfrac{6}{(x+3)^2}\Big)dx=5x-\dfrac{6}{x+3}+K$. 5. **Apply $R(0)=0$:** $0=0-\dfrac{6}{3}+K=-2+K\Rightarrow K=2$. 6. $R=5x-\dfrac{6}{x+3}+2$. …

The marginal revenue function for a firm is given by (5x²+30x+51)/((x+3)²). Show that the revenue function is

3.2 · Q2

\frac{5x^2+30x+51}{(x+3)^2}

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Concept understanding — Marginal Revenue Integration

Marginal Revenue Integration — From Intuition to Precision

Imagine you run a small bakery. You sell 10 cakes at ₹200 each — total revenue ₹2000. If you bake one more cake, you might have to lower the price to ₹190 to sell all 11. Your additional revenue from that 11th cake is not ₹190 — it's ₹190 minus the ₹10 you lost on each of the first 10 cakes (because you dropped the price). That net addition is your marginal revenue.

Marginal revenue (MR) is the change in total revenue when you sell one more unit. It answers: "What do I actually gain from the next sale, after accounting for any price change on existing units?"

The Intuition: Why MR Isn't Just Price

For a price-taker (perfect competition), MR equals price — selling one more unit doesn't force you to lower the price. But for any firm with market power (monopoly, monopolistic competition), selling more means lowering the price on all units. So MR is always less than price.

Think of it as: MR = Price of the new unit – Loss on previous units due to the price cut.

That loss is: (old price – new price) × (number of units sold before the cut).

The Precise Statement

Let total revenue $TR(q)$ be a function of quantity $q$ . Then marginal revenue at quantity $q$ is the derivative (or the discrete change):

MR(q) = \frac{d(TR)}{dq} \quad \text{or} \quad MR(q) = TR(q) - TR(q-1)

For a linear inverse demand curve $P = a - bq$ , total revenue is $TR = P \cdot q = aq - bq^2$ . Differentiate:

MR = \frac{d}{dq}(aq - bq^2) = a - 2bq

Notice: MR has the same intercept $a$ as demand, but twice the slope. This is the Marginal Revenue Integration idea — MR is derived from (integrated with) the demand curve.

MR = P + q \cdot \frac{dP}{dq}

Why "Integration"?

The term "marginal revenue integration" refers to the fact that total revenue is the integral (area under) the marginal revenue curve. Just as marginal cost integrates to total variable cost, marginal revenue integrates to total revenue: …

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Integrate $MR$ and fix the constant with $R(0)=0$ ; the result equals $\frac{2x}{x+3}+5x$ .

R(x)=\int MR\,dx

Given: $MR=\dfrac{5x^2+30x+51}{(x+3)^2}$ .
Rewrite the numerator about $(x+3)$ : $5x^2+30x+51=5(x^2+6x+9)+6=5(x+3)^2+6$ .
So $MR=\dfrac{5(x+3)^2+6}{(x+3)^2}=5+\dfrac{6}{(x+3)^2}$ .
Integrate: $R=\int\Big(5+\dfrac{6}{(x+3)^2}\Big)dx=5x-\dfrac{6}{x+3}+K$ .
Apply $R(0)=0$ : $0=0-\dfrac{6}{3}+K=-2+K\Rightarrow K=2$ .
$R=5x-\dfrac{6}{x+3}+2$ . …

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Q.The marginal revenue function for a firm is given by 5x2+30x+51(x+3)2\frac{5x^2+30x+51}{(x+3)^2}(x+3)25x2+30x+51​. Show that the revenue function is given by 2xx+3+5x\frac{2x}{x+3}+5xx+32x​+5x.

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Q.The marginal revenue function for a firm is given by $\frac{5x^2+30x+51}{(x+3)^2}$ . Show that the revenue function is given by $\frac{2x}{x+3}+5x$ .

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