How to calculate the indefinite integral of $(x-1)^{frac 12}-(x-3)$?
$begingroup$
Why is the following integral wrong?
$$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12(x-3)^2+c $$
The answer given by my textbook is:
$$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12x^2+3x+c $$
I understand why the textbook's answer is right. But I don't understand what I did wrong in mine.
calculus integration indefinite-integrals
edited Dec 9 '18 at 1:53
AccidentalFourierTransform
1,437827
asked Dec 8 '18 at 21:46
John ArgJohn Arg
316
$endgroup$
add a comment |
$begingroup$
Why is the following integral wrong?
$$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12(x-3)^2+c $$
The answer given by my textbook is:
$$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12x^2+3x+c $$
I understand why the textbook's answer is right. But I don't understand what I did wrong in mine.
calculus integration indefinite-integrals
edited Dec 9 '18 at 1:53
AccidentalFourierTransform
1,437827
asked Dec 8 '18 at 21:46
John ArgJohn Arg
316
$endgroup$
add a comment |
$begingroup$
Why is the following integral wrong?
$$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12(x-3)^2+c $$
The answer given by my textbook is:
$$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12x^2+3x+c $$
I understand why the textbook's answer is right. But I don't understand what I did wrong in mine.
calculus integration indefinite-integrals
edited Dec 9 '18 at 1:53
AccidentalFourierTransform
1,437827
asked Dec 8 '18 at 21:46
John ArgJohn Arg
316
$endgroup$
Why is the following integral wrong?
$$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12(x-3)^2+c $$
The answer given by my textbook is:
$$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12x^2+3x+c $$
I understand why the textbook's answer is right. But I don't understand what I did wrong in mine.
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Dec 9 '18 at 1:53
AccidentalFourierTransform
1,437827
asked Dec 8 '18 at 21:46
John ArgJohn Arg
316
edited Dec 9 '18 at 1:53
AccidentalFourierTransform
1,437827
asked Dec 8 '18 at 21:46
John ArgJohn Arg
316
edited Dec 9 '18 at 1:53
AccidentalFourierTransform
1,437827
edited Dec 9 '18 at 1:53
AccidentalFourierTransform
1,437827
edited Dec 9 '18 at 1:53
AccidentalFourierTransform
1,437827
1,437827
asked Dec 8 '18 at 21:46
John ArgJohn Arg
316
asked Dec 8 '18 at 21:46
John ArgJohn Arg
316
asked Dec 8 '18 at 21:46
John ArgJohn Arg
316
316
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that
begin{align}
frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x-3)^2+c &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x^2-6x+9)+c\ &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}x^2+3xunderbrace{-frac{9}{2}+c}_{text{constant}},
end{align}
so both answers are valid.
answered Dec 8 '18 at 21:48
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
begin{align}
frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x-3)^2+c &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x^2-6x+9)+c\ &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}x^2+3xunderbrace{-frac{9}{2}+c}_{text{constant}},
end{align}
so both answers are valid.
answered Dec 8 '18 at 21:48
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
$begingroup$
Note that
begin{align}
frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x-3)^2+c &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x^2-6x+9)+c\ &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}x^2+3xunderbrace{-frac{9}{2}+c}_{text{constant}},
end{align}
so both answers are valid.
answered Dec 8 '18 at 21:48
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
$begingroup$
Note that
begin{align}
frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x-3)^2+c &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x^2-6x+9)+c\ &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}x^2+3xunderbrace{-frac{9}{2}+c}_{text{constant}},
end{align}
so both answers are valid.
answered Dec 8 '18 at 21:48
MisterRiemannMisterRiemann
5,8451624
$endgroup$
Note that
begin{align}
frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x-3)^2+c &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x^2-6x+9)+c\ &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}x^2+3xunderbrace{-frac{9}{2}+c}_{text{constant}},
end{align}
so both answers are valid.
answered Dec 8 '18 at 21:48
MisterRiemannMisterRiemann
5,8451624
edited Dec 8 '18 at 21:54
answered Dec 8 '18 at 21:48
MisterRiemannMisterRiemann
5,8451624
answered Dec 8 '18 at 21:48
MisterRiemannMisterRiemann
5,8451624
answered Dec 8 '18 at 21:48
MisterRiemannMisterRiemann
5,8451624
5,8451624
add a comment |
add a comment |
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