a) Evaluate: x^2 4x + 2 .dx [3] b) If for 2 units of production, total revenue is Rs. 200, find the total revenue function given that MR = 10 + 20x 3x^2 . [3]

a) Evaluate: $\int \frac{x^2-4}{x+2}dx$

Step 1: Factorize the numerator

The integrand is $\frac{x^2-4}{x+2}$. First, factorize the numerator: $$x^2-4=(x+2)(x-2)$$ So the integrand becomes: $$\frac{(x+2)(x-2)}{x+2}$$

Step 2: Simplify the integrand

Cancel out the common factor $(x+2)$ (assuming $x\ne -2$): $$\frac{(x+2)(x-2)}{x+2}=x-2$$

Step 3: Integrate the simplified expression

Now, integrate the simplified expression: $$\int (x-2)dx=\int xdx-\int 2dx$$ $$=\frac{x^2}{2}-2x+C$$ where $C$ is the constant of integration.

Final Answer:

$$\boxed{\frac{x^2}{2}-2x+C}$$

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b) Find the total revenue function given:

• For 2 units of production, total revenue is Rs. 200.
• Marginal revenue function: $MR=10+20x-3x^2$.

Step 1: Recall the relationship between marginal revenue and total revenue

The marginal revenue (MR) is the derivative of the total revenue (TR) with respect to quantity $x$: $$MR=\frac{d(TR)}{dx}$$ Thus, to find the total revenue function, we integrate the marginal revenue function: $$TR=\int MRdx=\int (10+20x-3x^2)dx$$

Step 2: Integrate the marginal revenue function

$$TR=\int 10dx+\int 20xdx-\int 3x^{2}dx$$ $$=10x+10x^2-x^3+C$$ where $C$ is the constant of integration.

Step 3: Determine the constant of integration using the given condition

We are given that when $x=2$, $TR=200$. Substitute these values into the total revenue function: $$200=10(2)+10(2{)}^2-(2{)}^3+C$$ $$200=20+40-8+C$$ $$200=52+C$$ $$C=200-52=148$$

Step 4: Write the final total revenue function

Substitute $C=148$ back into the total revenue function: $$TR=10x+10x^2-x^3+148$$

Final Answer:

$$\boxed{TR=-x^3+10x^2+10x+148}$$