How do you find the definite integral for: #(x^43) (e^(-x^(44)) dx)# for the intervals #[0, 1]#?

Question

Christopher Allers · Accepted Answer

Use a #u#-substitution to get #int_0^1(x^43)(e^(-x^44))dx=(e-1)/(44e)~~0.0144#.

At first glance, this is a daunting problem. But it turns out the solution is found with a neat #u#-substitution. Note that, in #int_0^1(x^43)(e^(-x^44))dx#, we have a function (#x^44#) and its derivative (#x^43#). That makes it a perfect candidate for a #u#-substitution: Let #color(red)u=color(red)(x^44)->(du)/dx=44x^43->color(blue)(du)=color(blue)(44x^43dx)# Adjusting the integral to fit the substitution: #1/44int_0^1(color(blue)(44x^43))(e^(-color(red)(x^44)))color(blue)dx# Applying the substitution: #color(white)(XX)=1/44int_0^1e^(-color(red)(u))color(blue)(du)# This integral is now very simple: it's #-e^-u#: #1/44int_0^1e^(-color(red)(u))color(blue)(du)=1/44[-e^-u]_0^1# But wait! #u=x^44#, so #1/44int_0^1e^(-color(red)(u))color(blue)(du)=1/44[-e^-(x^44)]_0^1# Now we can evaluate: #1/44[-e^(x^44)]_0^1=1/44(-e^((1)^44)-(-e^((0)^44)))# #color(white)(XX)=1/44(-e^(-1)-(-e^(0)))# #color(white)(XX)=1/44(-1/e+1)# #color(white)(XX)=1/44((e-1)/e)# #color(white)(XX)=(e-1)/(44e)~~0.0144#

Axel Alvarez · Answer

undefined To find the definite integral of $ x^{43} e^{-x^{44}} $ over the interval [0, 1], you can use integration by substitution. Let $ u = -x^{44} $. Then, $ du = -44x^{43} dx $. Rearranging gives $ dx = -\frac{1}{44x^{43}} du $. Substitute these expressions into the integral:

$$ \int_{0}^{1} x^{43} e^{-x^{44}} dx = -\int_{0}^{-1} e^u du $$

This simplifies the integral to $ -\left[ e^u \right]_{0}^{-1} $, which becomes $ -(e^{-1} - e^0) $. Thus, the definite integral is $ e - 1 $.