What is #int_(0)^(2) xe^(x^2 + 2)dx #?

Answer 1

#1/2(e^6-e^2)#

which can be factored into

#e^2/2(e^2+1)(e^2-1)#

Approximately #198.02#

Perform a u-substitution

Let #u=x^2+2#
#(du)/dx=2x#
#dx=(du)/(2x)#
When #x=0#, #u=0^2+2=2#
When #x=2#, #u=2^2+2=6#

Now make the substitution into the integral

#int_2^6xe^u(du)/(2x)#
#1/2int_2^6e^udu#

Now integrate,

#1/2e^u#

Now evaluate

#1/2e^6-1/2e^2#
#1/2e^2(e^4-1)#
#1/2e^2(e^2+1)(e^2-1)#
This is approximately #198.02#
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Answer 2

To solve the integral ∫(0)^(2) xe^(x^2 + 2)dx, you can use integration by parts. Let u = x and dv = e^(x^2 + 2)dx. Then, differentiate u to get du = dx and integrate dv to get v = (1/2)e^(x^2 + 2). Now, apply the integration by parts formula: ∫udv = uv - ∫vdu.

Plugging in the values, you get: ∫(0)^(2) xe^(x^2 + 2)dx = [(1/2)x * e^(x^2 + 2)](0To solve the integral ∫(0)^(2) xe^(x^2 + 2)dx, you can use integration by parts. Let u = x and dv = e^(x^2 + 2)dx. Then, differentiate u to get du = dx and integrate dv to get v = (1/2)e^(x^2 + 2). Now, apply the integration by parts formula: ∫udv = uv - ∫vdu.

Plugging in the values, you get: ∫(0)^(2) xe^(x^2 + 2)dx = ^(2) - ∫(0)^(2) (1/2)e^(x^2 + 2)dx

Now, integrate the remaining part: = [(1/2)(2)e^(2 + 2)] - (1/2)e^(2 + 2) - [(1/2)e^(0 + 2)](2 - 0)

= e^4 - (1/2)e^4 - (1/2)e^2

= (1/2)e^4 - (1/2)e^2

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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.

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