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The function [tex]f[/tex] is continuous and [tex]\int_0^8 f(u) \, du = 6[/tex]. What is the value of [tex]\int_1^3 x f(x^2 - 1) \, dx[/tex]?

A. [tex]\frac{3}{2}[/tex]
B. 3
C. 6
D. 12
E. 24

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]

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