(e) Find all values of a Є R such that the integral is convergent. L₁ f(x)(1 − x²)ª dr ⇒ f(x) = We observe that f is continuous and odd in (-1, 1). It follows that it suffices to discuss the behavior of the integral in the interval [0, 1]. For x 1 we have x3 x3 f(x) = 2.3 x3 √√√x4 – 1| - x4 √(1 − x)(1 + x)(1 + x²) 1 √√1 − x√√(1 + x)(1 + x²) 2(1-x)1/2 Hence 1 f(x)(1 − x²), - 2(1−z)1/2(1−z)^2@ and the integral converges if 1½-½ - α < 1, that is a > −1, and it diverges if 29-1 (1 − x) ź—a - a 1 that is a ≤ -12.
(e) Find all values of a Є R such that the integral is convergent. L₁ f(x)(1 − x²)ª dr ⇒ f(x) = We observe that f is continuous and odd in (-1, 1). It follows that it suffices to discuss the behavior of the integral in the interval [0, 1]. For x 1 we have x3 x3 f(x) = 2.3 x3 √√√x4 – 1| - x4 √(1 − x)(1 + x)(1 + x²) 1 √√1 − x√√(1 + x)(1 + x²) 2(1-x)1/2 Hence 1 f(x)(1 − x²), - 2(1−z)1/2(1−z)^2@ and the integral converges if 1½-½ - α < 1, that is a > −1, and it diverges if 29-1 (1 − x) ź—a - a 1 that is a ≤ -12.
Chapter6: Exponential And Logarithmic Functions
Section6.7: Exponential And Logarithmic Models
Problem 27SE: Prove that bx=exln(b) for positive b1 .
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The question and the solution are in the attached photo
could you please explain the solution
![(e) Find all values of a Є R such that the integral
is convergent.
L₁ f(x)(1 − x²)ª dr
⇒ f(x)
=
We observe that f is continuous and odd in (-1, 1). It follows that it suffices to discuss the behavior of the integral in the
interval [0, 1]. For x 1 we have
x3
x3
f(x) =
2.3
x3
√√√x4 – 1|
- x4 √(1 − x)(1 + x)(1 + x²)
1
√√1 − x√√(1 + x)(1 + x²)
2(1-x)1/2
Hence
1
f(x)(1 − x²),
-
2(1−z)1/2(1−z)^2@
and the integral converges if 1½-½ - α < 1, that is a > −1, and it diverges if
29-1
(1 − x) ź—a
- a 1 that is a ≤ -12.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F5de84111-b761-4008-a23f-301da7a22077%2Fsw593z_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(e) Find all values of a Є R such that the integral
is convergent.
L₁ f(x)(1 − x²)ª dr
⇒ f(x)
=
We observe that f is continuous and odd in (-1, 1). It follows that it suffices to discuss the behavior of the integral in the
interval [0, 1]. For x 1 we have
x3
x3
f(x) =
2.3
x3
√√√x4 – 1|
- x4 √(1 − x)(1 + x)(1 + x²)
1
√√1 − x√√(1 + x)(1 + x²)
2(1-x)1/2
Hence
1
f(x)(1 − x²),
-
2(1−z)1/2(1−z)^2@
and the integral converges if 1½-½ - α < 1, that is a > −1, and it diverges if
29-1
(1 − x) ź—a
- a 1 that is a ≤ -12.
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