Evaluate the integrals in Exercises 31–78.
55. ∫(from -2...

Question

Evaluate the integrals in Exercises 31–78.

55. ∫(from -2 to -1)e^(-(x+1)) dx

Accepted Answer

Welcome back everyone. Evaluate the integral from -3 up to -2 of e to the power of x + 2 dx. Ae divided by 2, b e minus 1, C and D 1 minus e. We're going to approach this problem using U substitution. Let's suppose that U is equal to negative. Of x + 2 Our next step is to find the derivative of U with respect to x, which is going to be negative. We can factor out the negative sign. The derivative of x plus 2, which is basically -1 because the derivative of x is 1 and the derivative of a constant is 0. We can now show that DX is equal to DU divided by -1, which is DU. Now let's change the limits of integration. U1, the lower limit of integration, is going to be negative -3 + 2, which is negative of -1, and that's 1, and the upper limit of integration u2 is going to be negative of -2 + 2, which is 0. We can now rewrite the integral as the integral from 1 up to 0 of e to the power of u and dx becomes negative duu. So we can write d u and we can factor out the negative sign. Notice that this is an elementary integral. We get negative and the integral of e to the power of u duu is e to the power of u. We want to evaluate the result from 1. Up to 0. Now let's apply the fundamental theorem of calculus part 2. We get e to the power of 0 minus e to the power of 1. That's -1 + E. Or e minus 1, which corresponds to the answer choice b. Thank you for watching.

Evaluate the integrals in Exercises 31–78.55. ∫(from -2 to -1)e^(-(x+1)) dx

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Evaluate the integrals in Exercises 31–78.
55. ∫(from -2 to -1)e^(-(x+1)) dx