anonymous
  • anonymous
Calculas help? ∫??(5−ex)(e2x)dx
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits_{?}^{?}(5-e ^{x})(e ^{2x})\]
bahrom7893
  • bahrom7893
Multiply it out, it's gonna make your life much easier
bahrom7893
  • bahrom7893
Int (5e^(2x)-e^(3x),x)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
opps it is suppose to be divided by the second e
bahrom7893
  • bahrom7893
ohh that makes it a little different
anonymous
  • anonymous
yea i would think so lol
bahrom7893
  • bahrom7893
oh still rewrite it as: Int (5/e^(2x) - e^x/e^(2x))) Int (5*e^(-2x) - e^(-x))
bahrom7893
  • bahrom7893
Now split it into two: 5 * Int(e^(-2x)) - Int(e^(-x))
bahrom7893
  • bahrom7893
Now just use the rule: Int (e^(kx)) = e^(kx) / k
bahrom7893
  • bahrom7893
In the first half k = -2, in the second half, k = -1 So finally: 5 * (e^(-2x)/(-2)) + e^(-x) + C
anonymous
  • anonymous
ok that makes sense thanks
bahrom7893
  • bahrom7893
btw to prove that Int (e^(kx)) = e^(kx)/k just use u sub. Let u = kx, du = k dx, etc..

Looking for something else?

Not the answer you are looking for? Search for more explanations.