Mixing
Integral with respect to $x +$ constant
$begingroup$
Is this a valid expression:
$$int xd(x+5)$$
I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:
$$int (u - 5)du= frac{u^2}{2} - 5u + C = frac{(x+5)^2}{2} - 5(x+5) +C = frac{x^2}{2} -12.5 + C $$
Or is it correct to just do $d(x+5) = dx$ from the beginning and calculate $$int xd(x+5) = int xdx$$
calculus algebra-precalculus
asked Feb 1 at 5:29
SladeSlade
78111
$endgroup$
add a comment |
$begingroup$
Is this a valid expression:
$$int xd(x+5)$$
I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:
$$int (u - 5)du= frac{u^2}{2} - 5u + C = frac{(x+5)^2}{2} - 5(x+5) +C = frac{x^2}{2} -12.5 + C $$
Or is it correct to just do $d(x+5) = dx$ from the beginning and calculate $$int xd(x+5) = int xdx$$
calculus algebra-precalculus
asked Feb 1 at 5:29
SladeSlade
78111
$endgroup$
add a comment |
$begingroup$
Is this a valid expression:
$$int xd(x+5)$$
I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:
$$int (u - 5)du= frac{u^2}{2} - 5u + C = frac{(x+5)^2}{2} - 5(x+5) +C = frac{x^2}{2} -12.5 + C $$
Or is it correct to just do $d(x+5) = dx$ from the beginning and calculate $$int xd(x+5) = int xdx$$
calculus algebra-precalculus
asked Feb 1 at 5:29
SladeSlade
78111
$endgroup$
Is this a valid expression:
$$int xd(x+5)$$
I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:
$$int (u - 5)du= frac{u^2}{2} - 5u + C = frac{(x+5)^2}{2} - 5(x+5) +C = frac{x^2}{2} -12.5 + C $$
Or is it correct to just do $d(x+5) = dx$ from the beginning and calculate $$int xd(x+5) = int xdx$$
calculus algebra-precalculus
calculus algebra-precalculus
asked Feb 1 at 5:29
SladeSlade
78111
asked Feb 1 at 5:29
SladeSlade
78111
asked Feb 1 at 5:29
SladeSlade
78111
asked Feb 1 at 5:29
SladeSlade
78111
asked Feb 1 at 5:29
SladeSlade
78111
78111
add a comment |
add a comment |
2 Answers
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$begingroup$
Notice that you have $-12.5+C$, which is just another constant.
answered Feb 1 at 5:40
高田航高田航
1,365418
$endgroup$
$begingroup$
Oh, that makes a lot of sense now. Thanks!
$endgroup$
– Slade
Feb 1 at 5:59
add a comment |
$begingroup$
Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)
$$begin{align}
int xe^{-x^2}dx &= int frac{xe^{-x^2}dx}{1}\
&= int frac{xe^{-x^2}dx}{1}cdotfrac{frac{d(-x^2)}{dx}}{frac{d(-x^2)}{dx}}\
&= int frac{xe^{-x^2}dxcdotfrac{d(-x^2)}{dx}}{-2x}\
&= int frac{xe^{-x^2}d(-x^2)}{-2x}\
&= color{red}{-frac{1}{2}int e^{-x^2}d(-x^2)} &[1]\
&= color{green}{-frac{1}{2}int e^udu} &[2]\
&= -frac{1}{2}e^u + C\
&= -frac{1}{2}e^{-x^2} + C\
end{align}$$
answered Feb 2 at 12:56
John JoyJohn Joy
6,29911827
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Notice that you have $-12.5+C$, which is just another constant.
answered Feb 1 at 5:40
高田航高田航
1,365418
$endgroup$
$begingroup$
Oh, that makes a lot of sense now. Thanks!
$endgroup$
– Slade
Feb 1 at 5:59
add a comment |
$begingroup$
Notice that you have $-12.5+C$, which is just another constant.
answered Feb 1 at 5:40
高田航高田航
1,365418
$endgroup$
$begingroup$
Oh, that makes a lot of sense now. Thanks!
$endgroup$
– Slade
Feb 1 at 5:59
add a comment |
$begingroup$
Notice that you have $-12.5+C$, which is just another constant.
answered Feb 1 at 5:40
高田航高田航
1,365418
$endgroup$
Notice that you have $-12.5+C$, which is just another constant.
answered Feb 1 at 5:40
高田航高田航
1,365418
answered Feb 1 at 5:40
高田航高田航
1,365418
answered Feb 1 at 5:40
高田航高田航
1,365418
answered Feb 1 at 5:40
高田航高田航
1,365418
1,365418
$begingroup$
Oh, that makes a lot of sense now. Thanks!
$endgroup$
– Slade
Feb 1 at 5:59
add a comment |
$begingroup$
Oh, that makes a lot of sense now. Thanks!
$endgroup$
– Slade
Feb 1 at 5:59
$begingroup$
Oh, that makes a lot of sense now. Thanks!
$endgroup$
– Slade
Feb 1 at 5:59
$begingroup$
Oh, that makes a lot of sense now. Thanks!
$endgroup$
– Slade
Feb 1 at 5:59
add a comment |
$begingroup$
Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)
$$begin{align}
int xe^{-x^2}dx &= int frac{xe^{-x^2}dx}{1}\
&= int frac{xe^{-x^2}dx}{1}cdotfrac{frac{d(-x^2)}{dx}}{frac{d(-x^2)}{dx}}\
&= int frac{xe^{-x^2}dxcdotfrac{d(-x^2)}{dx}}{-2x}\
&= int frac{xe^{-x^2}d(-x^2)}{-2x}\
&= color{red}{-frac{1}{2}int e^{-x^2}d(-x^2)} &[1]\
&= color{green}{-frac{1}{2}int e^udu} &[2]\
&= -frac{1}{2}e^u + C\
&= -frac{1}{2}e^{-x^2} + C\
end{align}$$
answered Feb 2 at 12:56
John JoyJohn Joy
6,29911827
$endgroup$
add a comment |
$begingroup$
Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)
$$begin{align}
int xe^{-x^2}dx &= int frac{xe^{-x^2}dx}{1}\
&= int frac{xe^{-x^2}dx}{1}cdotfrac{frac{d(-x^2)}{dx}}{frac{d(-x^2)}{dx}}\
&= int frac{xe^{-x^2}dxcdotfrac{d(-x^2)}{dx}}{-2x}\
&= int frac{xe^{-x^2}d(-x^2)}{-2x}\
&= color{red}{-frac{1}{2}int e^{-x^2}d(-x^2)} &[1]\
&= color{green}{-frac{1}{2}int e^udu} &[2]\
&= -frac{1}{2}e^u + C\
&= -frac{1}{2}e^{-x^2} + C\
end{align}$$
answered Feb 2 at 12:56
John JoyJohn Joy
6,29911827
$endgroup$
add a comment |
$begingroup$
Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)
$$begin{align}
int xe^{-x^2}dx &= int frac{xe^{-x^2}dx}{1}\
&= int frac{xe^{-x^2}dx}{1}cdotfrac{frac{d(-x^2)}{dx}}{frac{d(-x^2)}{dx}}\
&= int frac{xe^{-x^2}dxcdotfrac{d(-x^2)}{dx}}{-2x}\
&= int frac{xe^{-x^2}d(-x^2)}{-2x}\
&= color{red}{-frac{1}{2}int e^{-x^2}d(-x^2)} &[1]\
&= color{green}{-frac{1}{2}int e^udu} &[2]\
&= -frac{1}{2}e^u + C\
&= -frac{1}{2}e^{-x^2} + C\
end{align}$$
answered Feb 2 at 12:56
John JoyJohn Joy
6,29911827
$endgroup$
Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)
$$begin{align}
int xe^{-x^2}dx &= int frac{xe^{-x^2}dx}{1}\
&= int frac{xe^{-x^2}dx}{1}cdotfrac{frac{d(-x^2)}{dx}}{frac{d(-x^2)}{dx}}\
&= int frac{xe^{-x^2}dxcdotfrac{d(-x^2)}{dx}}{-2x}\
&= int frac{xe^{-x^2}d(-x^2)}{-2x}\
&= color{red}{-frac{1}{2}int e^{-x^2}d(-x^2)} &[1]\
&= color{green}{-frac{1}{2}int e^udu} &[2]\
&= -frac{1}{2}e^u + C\
&= -frac{1}{2}e^{-x^2} + C\
end{align}$$
answered Feb 2 at 12:56
John JoyJohn Joy
6,29911827
edited Feb 2 at 13:05
answered Feb 2 at 12:56
John JoyJohn Joy
6,29911827
answered Feb 2 at 12:56
John JoyJohn Joy
6,29911827
answered Feb 2 at 12:56
John JoyJohn Joy
6,29911827
6,29911827
add a comment |
add a comment |
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