How do you determine if the improper integral converges or diverges #int 3/(7x^2 +4) # from negative infinity to infinity?

Answer 1

#int_(-oo)^(+oo) 3/(7x^2+4)dx = (3pi)/(2sqrt7)#

Given:

#f(x) = 3/(7x^2+4)#

Evaluate first:

#lim_(t->oo) int_0^t 3/(7x^2+4) dx = lim_(t->oo) 3/4 int_0^t dx/(7/4x^2+1) #
#lim_(t->oo) int_0^t 3/(7x^2+4) dx = lim_(t->oo) 3/(2sqrt7) int_0^t (d(sqrt7/2x))/( (sqrt7/2x)^2+1) #
#lim_(t->oo) int_0^t 3/(7x^2+4) dx = lim_(t->oo) 3/(2sqrt7) [arctan(sqrt7/2x)]_0^t #
#lim_(t->oo) int_0^t 3/(7x^2+4) dx = lim_(t->oo) 3/(2sqrt7) arctan(sqrt7/2t) #
#lim_(t->oo) int_0^t 3/(7x^2+4) dx = (3pi)/(4sqrt7)#

Then the integral:

#int_0^oo 3/(7x^2+4)dx = (3pi)/(4sqrt7)#

is convergent.

Evaluate then:

#int_(-t)^0 3/(7x^2+4)dx#
substitute now: #u=-x#:
#int_(-t)^0 3/(7x^2+4)dx = - int_t^0 3/(7(-u)^2+4)du = int_0^t 3/(7u^2+4)du#

so that also:

#int_(-oo)^0 3/(7x^2+4)dx = (3pi)/(4sqrt7)#

is convergent.

Finally then:

#int_(-oo)^(+oo) 3/(7x^2+4)dx = (3pi)/(2sqrt7)#
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Answer 2

To determine if the improper integral converges or diverges, we need to split it into two separate integrals and then evaluate each part individually.

The integral from negative infinity to infinity of the given function is equivalent to the sum of the integrals from negative infinity to 0 and from 0 to infinity.

For each of these integrals, we'll use the limit comparison test. We'll compare the given function to 1/x^2 as x approaches positive and negative infinity.

Let's evaluate each part:

  1. Integral from negative infinity to 0: As x approaches negative infinity, the given function behaves like 3/(7x^2). So, we'll compare it to 1/x^2. Limit comparison: lim (x -> -∞) (3/(7x^2 + 4)) / (1/x^2) Simplifying, we get: lim (x -> -∞) (3x^2) / (7x^2 + 4) This limit equals 3/7, which is a finite non-zero value.

  2. Integral from 0 to infinity: As x approaches infinity, the given function behaves like 3/(7x^2). So, we'll compare it to 1/x^2. Limit comparison: lim (x -> ∞) (3/(7x^2 + 4)) / (1/x^2) Simplifying, we get: lim (x -> ∞) (3x^2) / (7x^2 + 4) This limit also equals 3/7, a finite non-zero value.

Since both limits are finite non-zero values, the integral converges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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