Question

let ( f ) be a twice - differentiable function with first derivative ( f ) and second derivative ( f ). after applying integration by parts once to the integral (int 10x^{2}f(x^{2})dx) which of the following expressions could you obtain?
a. ( 10x^{2}f(x^{2})-int 20xf(x^{2})dx )
b. ( 5xf(x^{2})-int 5f(x^{2})dx )
c. ( 10x^{3}f(x^{3})-int 10f(x^{2})dx )
d. ( 5xf(x^{2})-int 5xf(x^{2})dx )
e. ( 5x^{3}f(x^{2})-int 10xf(x^{2})dx )

Explanation:

Step1: Recall Integration by Parts Formula

The integration by parts formula is \(\int u \, dv = uv - \int v \, du\). We need to choose \(u\) and \(dv\) for the integral \(\int 10x^{2}f'(x^{2})dx\). Let's make a substitution first to simplify. Let \(t = x^{2}\), then \(dt = 2xdx\), or \(xdx=\frac{dt}{2}\). But maybe it's easier to directly apply integration by parts. Let's set \(u = f'(x^{2})\) and \(dv = 10x^{2}dx\)? Wait, no, let's check the derivative of \(f(x^{2})\). The derivative of \(f(x^{2})\) with respect to \(x\) is \(f'(x^{2})\cdot 2x\) by the chain rule. So maybe we can rewrite the integral. Let's consider \(u = 5x\) and \(dv = 2xf'(x^{2})dx\)? Wait, let's compute \(du\) and \(v\) properly.

Wait, let's look at the integral \(\int 10x^{2}f'(x^{2})dx\). Let's rewrite \(10x^{2}dx\) as \(5x\cdot 2xdx\). Notice that \(2xdx\) is the derivative of \(x^{2}\), so if we let \(v = f(x^{2})\), then \(dv = f'(x^{2})\cdot 2xdx\). Then, let's set \(u = 5x\) and \(dv = 2xf'(x^{2})dx\). Wait, let's do it step by step.

Let’s set \(u = 5x\), then \(du = 5dx\). Wait, no, let's try \(u = 5x\) and \(dv = 2xf'(x^{2})dx\). Then \(v = f(x^{2})\) (since \(\int 2xf'(x^{2})dx = f(x^{2}) + C\), because the derivative of \(f(x^{2})\) is \(2xf'(x^{2})\)). Now, the integral \(\int 10x^{2}f'(x^{2})dx=\int 5x\cdot 2xf'(x^{2})dx\). So using integration by parts with \(u = 5x\) and \(dv = 2xf'(x^{2})dx\), we have:

\(\int u \, dv = uv - \int v \, du\)

Here, \(u = 5x\), \(dv = 2xf'(x^{2})dx\), so \(du = 5dx\), \(v = f(x^{2})\).

Step2: Apply Integration by Parts

Substitute into the formula:

\(\int 5x\cdot 2xf'(x^{2})dx=5x\cdot f(x^{2})-\int f(x^{2})\cdot 5dx\)

Simplify the left - hand side: \(\int 5x\cdot 2xf'(x^{2})dx=\int 10x^{2}f'(x^{2})dx\)

And the right - hand side: \(5xf(x^{2})-\int 5f(x^{2})dx\)

Answer:

B. \(5xf(x^{2})-\int 5f(x^{2})dx\)