How to calculate the indefinite integral of $(x-1)^{frac 12}-(x-3)$?
up vote
3
down vote
favorite
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 2 days ago
AccidentalFourierTransform
1,427827
asked 2 days ago
John Arg
266
add a comment |
up vote
3
down vote
favorite
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 2 days ago
AccidentalFourierTransform
1,427827
asked 2 days ago
John Arg
266
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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 2 days ago
AccidentalFourierTransform
1,427827
asked 2 days ago
John Arg
266
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 2 days ago
AccidentalFourierTransform
1,427827
asked 2 days ago
John Arg
266
edited 2 days ago
AccidentalFourierTransform
1,427827
asked 2 days ago
John Arg
266
edited 2 days ago
AccidentalFourierTransform
1,427827
edited 2 days ago
AccidentalFourierTransform
1,427827
edited 2 days ago
AccidentalFourierTransform
1,427827
1,427827
asked 2 days ago
John Arg
266
asked 2 days ago
John Arg
266
asked 2 days ago
John Arg
266
266
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
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 2 days ago
MisterRiemann
5,6691624
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
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 2 days ago
MisterRiemann
5,6691624
add a comment |
up vote
6
down vote
accepted
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 2 days ago
MisterRiemann
5,6691624
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
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 2 days ago
MisterRiemann
5,6691624
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 2 days ago
MisterRiemann
5,6691624
edited 2 days ago
answered 2 days ago
MisterRiemann
5,6691624
answered 2 days ago
MisterRiemann
5,6691624
answered 2 days ago
MisterRiemann
5,6691624
5,6691624
add a comment |
add a comment |
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