How do you find the definite integral of #2/(sqrt(x) *e^-sqrt(x))# from #[1, 4]#?

Question

Emma Aldridge · Accepted Answer

#4e^2-4e#

#I=int_1^4 2/(sqrtx*e^(-sqrtx))dx#

Simplify the negative exponent by bringing it to the numerator.

#I=2int_1^4e^sqrtx/sqrtxdx#

We will use the substitution #u=sqrtx#. This implies that #du=1/(2sqrtx)dx#. Don't forget to plug the bounds of #1# and #4# into #sqrtx#.

#I=4int_1^4e^sqrtx/(2sqrtx)dx=4int_1^2e^udu#

The integral #inte^udu=e^u+C#:

#I=4[e^u]_1^2=4(e^2-e^1)=4e^2-4e#

Scarlett Allison · Answer

undefined To find the definite integral of $ \frac{2}{\sqrt{x} \times e^{-\sqrt{x}}} $ from 1 to 4, you can use substitution. Let $ u = \sqrt{x} $, then $ du = \frac{1}{2\sqrt{x}} dx $. The integral becomes $ 4\int_{1}^{2} e^u du $. Integrating $ e^u $ with respect to $ u $ yields $ e^u $. Evaluate this from 1 to 2, which gives $ 4(e^2 - e) $ or approximately $ 35.692 $.

Charles Arocha · Answer

undefined To find the definite integral of $ \frac{2}{\sqrt{x} \times e^{-\sqrt{x}}} $ from $ x = 1 $ to $ x = 4 $, you can use the substitution method.

Let $ u = \sqrt{x} $. Then $ du = \frac{1}{2\sqrt{x}} dx $, which can be rewritten as $ 2\sqrt{x} \, du = dx $.

Now, substitute $ u $ and $ dx $ in terms of $ u $ into the integral:

$ \int_{1}^{4} \frac{2}{\sqrt{x} \times e^{-\sqrt{x}}} \, dx = \int_{u=1}^{u=2} \frac{2}{u \times e^{-u}} \times 2\sqrt{x} \, du $

This simplifies to:

$ \int_{1}^{2} \frac{4}{u \times e^{-u}} \, du $

Now, this integral can be solved as a standard integral. After integration, evaluate the result from $ u = 1 $ to $ u = 2 $ to find the definite integral.