Integral Calculus Function Transformation Trick

 

❓ Question

Let $f(x)$ be a polynomial function such that:

$$f(x^2+1)=x^4+5x^2+3$$

then:

$$\int_0^3 f(x)\,dx$$

is equal to:

Options:

1. $15$
   
2. $\dfrac{39}{2}$
3. $\dfrac{35}{2}$
4. $16$
   

---

🖼 Question Image

---

✍️ Short Explanation

This problem is based on:

👉 Function transformation
👉 Polynomial comparison
👉 Definite integration.

Main idea:

Convert:

$$f(x^2+1)$$

into standard polynomial form by substitution.

---

🔷 Step 1 — Put New Variable 💯

Given:

$$f(x^2+1)=x^4+5x^2+3$$

Let:

$$t=x^2+1$$

Then:

$$x^2=t-1$$

and:

$$x^4=(t-1)^2$$

Substitute into RHS:

$$f(t) = (t-1)^2+5(t-1)+3$$

---

🔷 Step 2 — Simplify Polynomial

Expand:

$$(t-1)^2=t^2-2t+1$$

Thus:

$$f(t) = t^2-2t+1+5t-5+3$$
$$=t^2+3t-1$$

Hence:

$$\boxed{ f(x)=x^2+3x-1 }$$

---

🔷 Step 3 — Evaluate Integral

Now:

$$\int_0^3 f(x)\,dx = \int_0^3 (x^2+3x-1)\,dx$$

Integrate:

$$= \left[ \frac{x^3}{3} +\frac{3x^2}{2} -x \right]_0^3$$

Substitute limits:

$$= \left( \frac{27}{3} +\frac{3(9)}{2} -3 \right)$$
$$= 9+\frac{27}{2}-3$$
$$= 6+\frac{27}{2}$$
$$= \frac{12+27}{2}$$
$$= \frac{39}{2}$$

---

🔷 Step 4 — JEE Trap Alert 🚨

❌ Directly $x^2+1=x$ assume kar lena

❌ Polynomial expansion me sign mistake kar dena

❌ Final integration calculation me error karna

Remember:

$$\boxed{ x^2=t-1 }$$

is the key substitution.

---

✅ Final Answer

$$\boxed{ \frac{39}{2} }$$

(Option 2)

---

📚 Related Topics

• Find Polynomial Value Using Root Formation Method
• Find Sum of Fourth Powers Using Symmetric Identities
• Shortest Distance Between Lines and Ellipse Relation