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Answer :
To solve the problem, we want to evaluate the integral [tex]\(\int_1^3 x f(x^2 - 1) \, dx\)[/tex] given that [tex]\(\int_0^8 f(u) \, du = 6\)[/tex].
We'll use the method of substitution to evaluate this integral. Let's follow these steps:
1. Substitution:
- Let [tex]\( u = x^2 - 1 \)[/tex]. This substitution will help simplify the integral because it matches the argument of the function [tex]\( f \)[/tex].
- Then, calculate the derivative: [tex]\( \frac{du}{dx} = 2x \)[/tex], which implies [tex]\( du = 2x \, dx \)[/tex] or [tex]\( dx = \frac{du}{2x} \)[/tex].
2. Adjust limits of integration:
- Change the limits of [tex]\( x \)[/tex] to limits in terms of [tex]\( u \)[/tex]:
- When [tex]\( x = 1 \)[/tex], [tex]\( u = 1^2 - 1 = 0 \)[/tex].
- When [tex]\( x = 3 \)[/tex], [tex]\( u = 3^2 - 1 = 8 \)[/tex].
- So, the limits for [tex]\( u \)[/tex] are from 0 to 8.
3. Rewrite the integral:
- Substitute [tex]\( u \)[/tex] and [tex]\( dx \)[/tex] into the integral:
[tex]\[
\int_1^3 x f(x^2 - 1) \, dx = \int_0^8 f(u) \cdot \frac{du}{2x}
\][/tex]
- Notice that [tex]\( 2x \)[/tex] in the denominator cancels out with [tex]\( x \)[/tex] in [tex]\( x \cdot \frac{du}{2x} \)[/tex]. So:
[tex]\[
\int_1^3 x f(x^2 - 1) \, dx = \int_0^8 \frac{1}{2} f(u) \, du
\][/tex]
4. Evaluate the new integral:
- The integral [tex]\(\int_0^8 \frac{1}{2} f(u) \, du\)[/tex] simplifies to [tex]\(\frac{1}{2} \times \int_0^8 f(u) \, du\)[/tex].
- We know [tex]\(\int_0^8 f(u) \, du = 6\)[/tex].
5. Calculate the result:
- Multiply and simplify:
[tex]\[
\frac{1}{2} \times 6 = 3
\][/tex]
Therefore, the value of [tex]\(\int_1^3 x f(x^2 - 1) \, dx\)[/tex] is [tex]\(\boxed{3}\)[/tex].
We'll use the method of substitution to evaluate this integral. Let's follow these steps:
1. Substitution:
- Let [tex]\( u = x^2 - 1 \)[/tex]. This substitution will help simplify the integral because it matches the argument of the function [tex]\( f \)[/tex].
- Then, calculate the derivative: [tex]\( \frac{du}{dx} = 2x \)[/tex], which implies [tex]\( du = 2x \, dx \)[/tex] or [tex]\( dx = \frac{du}{2x} \)[/tex].
2. Adjust limits of integration:
- Change the limits of [tex]\( x \)[/tex] to limits in terms of [tex]\( u \)[/tex]:
- When [tex]\( x = 1 \)[/tex], [tex]\( u = 1^2 - 1 = 0 \)[/tex].
- When [tex]\( x = 3 \)[/tex], [tex]\( u = 3^2 - 1 = 8 \)[/tex].
- So, the limits for [tex]\( u \)[/tex] are from 0 to 8.
3. Rewrite the integral:
- Substitute [tex]\( u \)[/tex] and [tex]\( dx \)[/tex] into the integral:
[tex]\[
\int_1^3 x f(x^2 - 1) \, dx = \int_0^8 f(u) \cdot \frac{du}{2x}
\][/tex]
- Notice that [tex]\( 2x \)[/tex] in the denominator cancels out with [tex]\( x \)[/tex] in [tex]\( x \cdot \frac{du}{2x} \)[/tex]. So:
[tex]\[
\int_1^3 x f(x^2 - 1) \, dx = \int_0^8 \frac{1}{2} f(u) \, du
\][/tex]
4. Evaluate the new integral:
- The integral [tex]\(\int_0^8 \frac{1}{2} f(u) \, du\)[/tex] simplifies to [tex]\(\frac{1}{2} \times \int_0^8 f(u) \, du\)[/tex].
- We know [tex]\(\int_0^8 f(u) \, du = 6\)[/tex].
5. Calculate the result:
- Multiply and simplify:
[tex]\[
\frac{1}{2} \times 6 = 3
\][/tex]
Therefore, the value of [tex]\(\int_1^3 x f(x^2 - 1) \, dx\)[/tex] is [tex]\(\boxed{3}\)[/tex].
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