29–62. Integrals Evaluate the following integrals. Include abs...

Question

29–62. Integrals Evaluate the following integrals. Include absolute values only when needed.

∫₋₂² (e^{z/2}) / (e^{z/2} + 1) dz

Accepted Answer

Welcome back, everyone. Evaluate the integral from -1 to 3 of E raises the power of X divided by 2, divided by E raises the power of X divided by 2 + 1. For this problem, we're going to use U substitution. Let's suppose that U is equal to the denominator E raises the power of X divided by 2 + 1. The next step is to find the derivative of you with respect to X. The derivative of e raised the power of X. Divided by 2 is E raise the power of X divided by 2, and according to the chain rule, we're multiplying by the derivative of X divided by 2, which is 1/2. The derivative of 1 is 0. From this relationship we can show that our numerator. Of the integrat, which is eases the power of x divided by 2 multiplied by D X is equal to DU divided by 12 or simply to DU. We're now going to find our limits of integration in terms of U. U1, the lower limit of integration is going to be E. Raise the power of -12 + 1, and the upper limit of integration U2 is going to be equal to E. Raise the power of 3 halves plus 1. So that we can show that our integral becomes integral from. E raises the power of -12 + 1 up to E raises the power of three halves plus 1. Our numerator is 2DU. We can factor out 2. And we can write the u in the numerator and our denominator is simply U. Now, let's evaluate the result. We can notice that this is a standard integral. So we get 2 multiplied by LM of the absolute value of you. Now, do we need to use the absolute value? Well, the answer is no, because our limits of integration are positive numbers, so we can drop the absolute value and evaluate 2 LN of U from the lower limit of integration E to the power of -12 + 1 up to E raise to the power of 3 halves plus 1. Applying the fundamental theorem of calculus, part two, we get 2 impreis Ln of E raises the power of three halves plus 1 minus Ln of E raises the power of -1/2 + 1. And now let's go ahead and. Simplify Using the properties of logarithms, we get 2 LN imprecs. Our numerator contains e raises the power of three halves plus 1. And we're dividing by E, a race to the power of -1/2 + 1. Well then, that will be our final answer and thank you for watching.

29–62. Integrals Evaluate the following integrals. Include absolute values only when needed.
∫₋₂² (e^{z/2}) / (e^{z/2} + 1) dz

Key Concepts

Definite Integrals and Symmetry

Substitution Method

Exponential Functions and Their Properties

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