Answer :
We start with the integral
[tex]$$
\int_1^3 x\, f\left(x^2-1\right) \, dx.
$$[/tex]
A good strategy for this problem is to make a substitution that simplifies the argument of the function [tex]$f$[/tex]. Let
[tex]$$
u = x^2 - 1.
$$[/tex]
Then, differentiate both sides with respect to [tex]$x$[/tex]:
[tex]$$
\frac{du}{dx} = 2x \quad \Longrightarrow \quad du = 2x\,dx.
$$[/tex]
This implies that
[tex]$$
x\,dx = \frac{du}{2}.
$$[/tex]
Next, we change the limits of integration according to our substitution. When [tex]$x = 1$[/tex]:
[tex]$$
u = 1^2 - 1 = 0,
$$[/tex]
and when [tex]$x = 3$[/tex]:
[tex]$$
u = 3^2 - 1 = 9 - 1 = 8.
$$[/tex]
Now rewrite the integral in terms of [tex]$u$[/tex]:
[tex]$$
\int_1^3 x\, f\left(x^2-1\right) dx = \int_0^8 f(u) \frac{du}{2} = \frac{1}{2} \int_0^8 f(u) \, du.
$$[/tex]
We are given that
[tex]$$
\int_0^8 f(u) \, du = 6.
$$[/tex]
Thus, substituting this value in gives
[tex]$$
\frac{1}{2} \int_0^8 f(u) \, du = \frac{1}{2} \times 6 = 3.
$$[/tex]
Therefore, the value of the integral is
[tex]$$
\boxed{3}.
$$[/tex]
[tex]$$
\int_1^3 x\, f\left(x^2-1\right) \, dx.
$$[/tex]
A good strategy for this problem is to make a substitution that simplifies the argument of the function [tex]$f$[/tex]. Let
[tex]$$
u = x^2 - 1.
$$[/tex]
Then, differentiate both sides with respect to [tex]$x$[/tex]:
[tex]$$
\frac{du}{dx} = 2x \quad \Longrightarrow \quad du = 2x\,dx.
$$[/tex]
This implies that
[tex]$$
x\,dx = \frac{du}{2}.
$$[/tex]
Next, we change the limits of integration according to our substitution. When [tex]$x = 1$[/tex]:
[tex]$$
u = 1^2 - 1 = 0,
$$[/tex]
and when [tex]$x = 3$[/tex]:
[tex]$$
u = 3^2 - 1 = 9 - 1 = 8.
$$[/tex]
Now rewrite the integral in terms of [tex]$u$[/tex]:
[tex]$$
\int_1^3 x\, f\left(x^2-1\right) dx = \int_0^8 f(u) \frac{du}{2} = \frac{1}{2} \int_0^8 f(u) \, du.
$$[/tex]
We are given that
[tex]$$
\int_0^8 f(u) \, du = 6.
$$[/tex]
Thus, substituting this value in gives
[tex]$$
\frac{1}{2} \int_0^8 f(u) \, du = \frac{1}{2} \times 6 = 3.
$$[/tex]
Therefore, the value of the integral is
[tex]$$
\boxed{3}.
$$[/tex]
