Error in taking the derivative of an integral

[jesuslovesu]

I know that it's 6x^2 - 2 but when I'm trying take the derivative of the integral shouldn't I have to multiply each term by -1 because the x is in the lower bound? It gives a wrong answer, so am I doing something wrong or is it just that I'm not supposed to take the opposite in this case?

$$\[ \int_x^{-1} (2-6t^2)\,dt\] -1 * 2(-1 - x) - -1*6*(-1-x^3)/3$$

 

[topsquark]

I know that it's 6x^2 - 2 but when I'm trying take the derivative of the integral shouldn't I have to multiply each term by -1 because the x is in the lower bound? It gives a wrong answer, so am I doing something wrong or is it just that I'm not supposed to take the opposite in this case?

$$\[ \int_x^{-1} (2-6t^2)\,dt\] -1 * 2(-1 - t) - -1*6*(-1-t^3)/3$$

$$\int_x^{-1}(2-6t^2) dt$$
$$(2t -2t^3)|_x^{-1}$$

 

[topsquark]

I know that it's 6x^2 - 2 but when I'm trying take the derivative of the integral shouldn't I have to multiply each term by -1 because the x is in the lower bound? It gives a wrong answer, so am I doing something wrong or is it just that I'm not supposed to take the opposite in this case?

$$\[ \int_x^{-1} (2-6t^2)\,dt\] -1 * 2(-1 - t) - -1*6*(-1-t^3)/3$$

$$\int_x^{-1}(2-6t^2) dt$$
$$(2t -2t^3)|_x^{-1}$$
$$[2(-1)-2(-1)^3]-[2x-2x^3]$$
$$2x^3-2x$$

-Dan

 

[jesuslovesu]

In this case unfortunately I have to do it the long way using

c*(b-a) and (b^3-a^3)/3

 

[topsquark]

In this case unfortunately I have to do it the long way using

c*(b-a) and (b^3-a^3)/3

It's the same thing, just rearrange the terms:

$$(2t -2t^3)|_x^{-1} = 2(-1-x)-2[(-1)^3-x^3]$$

-Dan

 

[VietDao29]

$$\int_x^{-1}(2-6t^2) dt$$
$$(2t -2t^3)|_x^{-1}$$
$$[2(-1)-2(-1)^3]-[2x-2x^3]$$
$$2x^3-2x$$

-Dan

May I suggest you NOT to give out COMPLETE solutions? :grumpy: :grumpy: :grumpy: :grumpy: