Evaluating Integral ∫(1+2x³) dx from 0 to 5 for Answer 635/2

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Homework Statement


[tex]\int[/tex][tex]^{5}_{0}[/tex] 1+2x[tex]^{3}[/tex]


Homework Equations



answer is: 635/2

The Attempt at a Solution


Integrating the function I get this: 1/2(x+x[tex]^{}4[/tex])
My answer when evaluating the limits =1/2(630)
 
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jimen113 said:

Homework Statement


[tex]\int[/tex][tex]^{5}_{0}[/tex] 1+2x[tex]^{3}[/tex]


Homework Equations



answer is: 635/2

The Attempt at a Solution


Integrating the function I get this: 1/2(x+x[tex]^{}4[/tex])
You might want to re-check your integral. What is:

[tex]\int 1 dx[/tex]
 
[tex]\int1[/tex]=x
[tex]\int 2x^3[/tex] = [tex]\frac{x^4}{2}[/tex],
1/2[tex]\int[/tex]x+x^4
I took (1/2) out of the [tex]\frac{X^{4}}{2}[/tex]
(So, maybe I can't do that, I should leave it and evaluate at the limits using (x^4/2)?
 
jimen113 said:
[tex]\int1[/tex]=x
[tex]\int 2x^3[/tex] = [tex]\frac{x^4}{2}[/tex],
1/2[tex]\int[/tex]x+x^4
I took (1/2) out of the [tex]\frac{X^{4}}{2}[/tex]
(So, maybe I can't do that, I should leave it and evaluate at the limits using (x^4/2)?
Note that:

[tex]x+\frac{1}{2}x^4 \neq \frac{1}{2}\left(x+x^4\right)[/tex]

So yes, you need to evaluate:

[tex]\left.\left(x+\frac{1}{2}x^4\right)\right|_0^5[/tex]
 
I see where I went wrong, thank you for your help!