How do you evaluate the definite integral #int (2)/(2x+5) * dx# from #[1, e]#?

Answer 1

Use substitution.

#int_1^e (2)/(2x+5) * dx = int_1^e 1/(2x+5) 2 dx#
Let #u = 2x+5# so #du = 2dx# and
when #x=1#, #u=7# and
when #x=e#, #u = 2e+5#
#int_1^e 1/(2x+5) 2 dx = int_7^(2e+5) 1/u du#
# = lnu]_7^(2e+5)#
# = ln(2e+5)-ln7#
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Answer 2

To evaluate the definite integral ∫(2)/(2x+5) dx from 1 to e:

  1. Find the antiderivative of the function (2)/(2x+5) with respect to x, which is ln|2x+5| + C.
  2. Evaluate the antiderivative at the upper limit of integration (e) and subtract the value of the antiderivative at the lower limit of integration (1).
  3. Substitute the upper limit (e) into the antiderivative: ln|2(e)+5| = ln|2e+5|.
  4. Substitute the lower limit (1) into the antiderivative: ln|2(1)+5| = ln|7|.
  5. Subtract the value at the lower limit from the value at the upper limit: ln|2e+5| - ln|7|.
  6. Simplify: ln| (2e+5) / 7 |. This is the value of the definite integral.
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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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