What is #f(x) = int (2x-xe^x)(-xe^x+secx) dx# if #f(0 ) = 5 #?

Answer 1

Can't be perfectly determined.

First, Integrate.

So, #int (2x - xe^x)(-xe^x + secx)dx#
#= int (-2x^2e^x -x^2e^(2x) + 2xsec x - xe^xsecx)dx#
#= -2int x^2e^xdx -int x^2e^(2x)dx +2 int xsecx dx -int xe^xsecxdx#

Now We have to Integrate By Parts.

Assume #I_1 = -2 int x^2e^xdx#,
#I_2 = -int x^2e^(2x)dx#
#I_3 = 2int x sec xdx#
#I_4 = -intxe^xsecxdx#

Now The Problem is,

#I_1# and #I_2# can be found, but #I_3# and #I_4# can't be expressed by elementary functions. These integrals are undeterminable.
So, Get the #I_1# first.
#I_1 = -2[x^2inte^xdx - int {d/dx(x^2)inte^x dx}dx]#
#= -2 [x^2e^x - 2 intxe^xdx]#
#= -2[x^2e^x - 2{x inte^xdx - int (d/dx(x) int e^xdx)dx}]#
#= -2[x^2e^x - 2{xe^x - inte^xdx}]#
#= -2[x^2e^x - 2{xe^x -e^x}]#
#= -2[x^2e^x - 2xe^x +2e^x]#
#= -2e^x[x^2 - 2x + 2]#
Now, #I_2#.
#I_2 = -2[x^2inte^(2x)dx - int {d/dx(x^2)inte^(2x) dx}dx]#
#= -2 [1/2x^2e^(2x) - 2 intxe^(2x)dx]#
#= -2[1/2x^2e^(2x) - 2{x inte^(2x)dx - int (d/dx(x) int e^(2x)dx)dx}]#
#= -2[1/2x^2e^(2x) - 2{1/2xe^(2x) - inte^(2x)dx}]#
#= -2[1/2x^2e^(2x) - 2{1/2xe^(2x) -1/2e^(2x)}]#
#= -2[1/2x^2e^(2x) - xe^(2x) +e^(2x)]#
#= -2e^(2x)[1/2x^2 - x + 1]#

So, The Entire Integral is :

#int(2x - xe^x)(-xe^x + sec x) dx#
#= -2e^x[x^2 - 2x + 2] -2e^(2x)[1/2x^2 - x + 1] + 2 intxsecxdx - intxe^xsecxdx#

Hope this helps, but It won't, most probably.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find ( f(x) ), we need to integrate the given function ( (2x - xe^x)(-xe^x + \sec(x)) ) with respect to ( x ). Then, we'll use the initial condition ( f(0) = 5 ) to determine the constant of integration.

[ \begin{align*} f(x) &= \int (2x - xe^x)(-xe^x + \sec(x)) , dx \ &= \int (-2x^2e^x + xe^{2x} - 2x\sec(x) + xe^x\sec(x)) , dx \ &= -\frac{2}{3}x^3e^x + \frac{1}{2}xe^{2x} - 2x\ln|\sec(x) + \tan(x)| + \frac{1}{2}x^2 + C \end{align*} ]

Given that ( f(0) = 5 ), we can substitute ( x = 0 ) into the expression for ( f(x) ) and solve for ( C ).

[ \begin{align*} 5 &= -\frac{2}{3}(0)^3e^0 + \frac{1}{2}(0)e^{2 \cdot 0} - 2(0)\ln|\sec(0) + \tan(0)| + \frac{1}{2}(0)^2 + C \ 5 &= 0 + 0 - 0 + 0 + C \ C &= 5 \end{align*} ]

Therefore, the function ( f(x) ) is:

[ f(x) = -\frac{2}{3}x^3e^x + \frac{1}{2}xe^{2x} - 2x\ln|\sec(x) + \tan(x)| + \frac{1}{2}x^2 + 5 ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7