Evaluate the integrals in Exercises 87–96.
89. ∫₀¹ 2^(−θ...

Question

Evaluate the integrals in Exercises 87–96.

89. ∫₀¹ 2^(−θ) dθ

Accepted Answer

Welcome back everyone. Evaluate the integral from 0 to 1 of 3 to the power of negative x dx. A 1 divided by ln of 3, B 1 divided by 3 ln of 3. C 2 divided by ln of 3, and D 2 divided by 3 ln of 3. For this problem, let's begin by rewriting the integral as the integral from 0 to 1 of 1/3 raise to the power of x dx, right? Since we have a negative exponent, we can basically apply the reciprocal rule. We can flip 3 into 1/3, and we can turn our exponent into a positive one. And the reason why this is helpful is basically because we know that the integral of a to the power of x d x is equal to a power of x divided by ln of a plus a constant of integration c. So by analogy we're going to get a to the power of x, which is now 1/3. Raise the power of x. Divided by ln of 1/3, we do not have to add a constant of integration because we have a definite integral. We just want to evaluate the result from 0 to 1. Before we do that, we're going to simplify slightly, right, and show that this is going to be equal to. 1/3 raised to the power of x and ln of 1/3 can be written as negative ln of 3 using the properties of logarithms. We want to find the result from 0 to 1, we can factor out the constant which is 1 divided by ln of 3. And in parentheses we can take 1/3. Raise the power of 1 minus. 1/3 raised the power of 0. From here we are going to simplify. In the numerator we're going to have 1/3 minus 1. We also have that negative sign in front and in the denominator that's ln of 3. Now 1/3 minus 1 is 2/3, and 2/3 is 2/3. Which is divided by Alan of 3. And now combining the two denominators, we get 2 divided by 3, a lot of 3. Which corresponds to the answer to the Thank you for watching.

Evaluate the integrals in Exercises 87–96.89. ∫₀¹ 2^(−θ) dθ

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Evaluate the integrals in Exercises 87–96.
89. ∫₀¹ 2^(−θ) dθ