Question

If \(\int f(x) dx = F(x)\), then \(\int x^3 f(x^2) dx\) is equal to

• A. \(\frac{1}{2}[x^2\{F(x)\}^2 - \int \{F(x)\}^2 dx]\)
• B. \(\frac{1}{2}[x^2 F(x^2) - \int F(x^2) d(x^2)]\)
• C. \(\frac{1}{2}[x^2 F(x) - \frac{1}{2}\int \{F(x)\}^2 dx]\)
• D. None of the above

Hint:

Use substitution \(t = x^2\) followed by integration by parts.

Answer 1

The Correct Option is B

Let's evaluate the integral \(\int x^3 f(x^2) \, dx\). Given that \(\int f(x) \, dx = F(x)\), we can use substitution and integration techniques to solve this.

1. First, use substitution to simplify the integral. Let \(u = x^2\). This gives us \(du = 2x \, dx\) or \(x \, dx = \frac{1}{2} du\).
2. Substitute into the integral: \(\int x^3 f(x^2) \, dx = \int x^2 \cdot x f(x^2) \, dx\). Since \(x^2 = u\) and \(x \, dx = \frac{1}{2} du\), this becomes: \(\int u \cdot \frac{1}{2} f(u) \, du = \frac{1}{2} \int u f(u) \, du\).
3. Now, use integration by parts on \(\int u f(u) \, du\). Let \(v = F(u)\), the antiderivative of \(f(u)\), and \(dv = f(u) \, du\). Let \(w = u\) and \(dw = du\). According to integration by parts \(\int v \, dw = vw - \int w \, dv\), we have:
   • \(v = F(u)\)
   • \(w = u\)
   • \(dw = du\)
   • \(dv = f(u) \, du\)
4. Substituting back \(u = x^2\), the integral becomes: \(\frac{1}{2} \left[ x^2 F(x^2) - \int F(x^2) \, d(x^2) \right]\).
5. This matches the option:

\(\frac{1}{2} [x^2 F(x^2) - \int F(x^2) \, d(x^2)]\)

Thus, the correct answer is: \(\frac{1}{2} [x^2 F(x^2) - \int F(x^2) \, d(x^2)]\).