What is #f(x) = int -e^(2x)-x dx# if #f(-3 ) = 1 #?

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Answer 1 · 2017-05-12T19:22:13

# f(x) = -1/2e^(2x) - 1/2x^2 + 11/2+1/2e^(-6) #

We have:

# f(x) = int \ -e^(2x)-x \ dx #

We can integrate this directly to get:

# f(x) = -1/2e^(2x) - 1/2x^2+ C #

We are given that #f(-3)=1#

# :. -1/2e^(-6) - 1/2 9 + C = 1#
# :. C = 11/2+1/2e^(-6) #

Leading to:

# f(x) = -1/2e^(2x) - 1/2x^2 + 11/2+1/2e^(-6) #