Integral $int_0^1 frac{sqrt{x}}{3+x}; dx$ [closed]
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How can we compute integration below?
$$int_0^1 frac{sqrt{x}}{3+x} ;dx$$
Solution shows the answer is 2-π/3 but I can not derive this equation.
integration
edited Nov 15 at 20:34
MPW
29.4k11856
asked Nov 15 at 20:29
owen
91
closed as off-topic by Did, Gibbs, Leucippus, Scientifica, Anurag A Nov 16 at 3:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-2
down vote
favorite
How can we compute integration below?
$$int_0^1 frac{sqrt{x}}{3+x} ;dx$$
Solution shows the answer is 2-π/3 but I can not derive this equation.
integration
edited Nov 15 at 20:34
MPW
29.4k11856
asked Nov 15 at 20:29
owen
91
closed as off-topic by Did, Gibbs, Leucippus, Scientifica, Anurag A Nov 16 at 3:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Gibbs, Leucippus, Scientifica, Anurag A
If this question can be reworded to fit the rules in the help center, please edit the question.
HINT: Change of variables $x=t^2$
– Tito Eliatron
Nov 15 at 20:34
And even, $$x=3t^2$$
– Did
Nov 15 at 20:36
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
How can we compute integration below?
$$int_0^1 frac{sqrt{x}}{3+x} ;dx$$
Solution shows the answer is 2-π/3 but I can not derive this equation.
integration
edited Nov 15 at 20:34
MPW
29.4k11856
asked Nov 15 at 20:29
owen
91
How can we compute integration below?
$$int_0^1 frac{sqrt{x}}{3+x} ;dx$$
Solution shows the answer is 2-π/3 but I can not derive this equation.
integration
integration
edited Nov 15 at 20:34
MPW
29.4k11856
asked Nov 15 at 20:29
owen
91
edited Nov 15 at 20:34
MPW
29.4k11856
asked Nov 15 at 20:29
owen
91
edited Nov 15 at 20:34
MPW
29.4k11856
edited Nov 15 at 20:34
MPW
29.4k11856
edited Nov 15 at 20:34
MPW
29.4k11856
29.4k11856
asked Nov 15 at 20:29
owen
91
asked Nov 15 at 20:29
owen
91
asked Nov 15 at 20:29
owen
91
91
closed as off-topic by Did, Gibbs, Leucippus, Scientifica, Anurag A Nov 16 at 3:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Gibbs, Leucippus, Scientifica, Anurag A
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Gibbs, Leucippus, Scientifica, Anurag A Nov 16 at 3:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Gibbs, Leucippus, Scientifica, Anurag A
If this question can be reworded to fit the rules in the help center, please edit the question.
HINT: Change of variables $x=t^2$
– Tito Eliatron
Nov 15 at 20:34
And even, $$x=3t^2$$
– Did
Nov 15 at 20:36
add a comment |
HINT: Change of variables $x=t^2$
– Tito Eliatron
Nov 15 at 20:34
And even, $$x=3t^2$$
– Did
Nov 15 at 20:36
HINT: Change of variables $x=t^2$
– Tito Eliatron
Nov 15 at 20:34
HINT: Change of variables $x=t^2$
– Tito Eliatron
Nov 15 at 20:34
And even, $$x=3t^2$$
– Did
Nov 15 at 20:36
And even, $$x=3t^2$$
– Did
Nov 15 at 20:36
add a comment |
2 Answers
2
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1
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accepted
Make the substitution: $x=u^2, , dx=2u,du$
$$int_0^1 frac{sqrt{x}}{3+x} dx=2int_0^1 frac{u^2}{3+u^2} du={displaystyleint_0^1}left(dfrac{class{steps-node}{cssId{steps-node-4}{u^2+3}}}{u^2+3}-dfrac{class{steps-node}{cssId{steps-node-5}{3}}}{u^2+3}right)mathrm{d}u={displaystyleint_0^1}left(1-dfrac{3}{u^2+3}right)mathrm{d}u$$
$$={displaystyleint_0^1}1,mathrm{d}u-class{steps-node}{cssId{steps-node-6}{3}}{displaystyleint_0^1}dfrac{1}{u^2+3},mathrm{d}u=1-class{steps-node}{cssId{steps-node-7}{dfrac{1}{sqrt{3}}}}{displaystyleint_0^1}dfrac{1}{v^2+1},mathrm{d}v=1-sqrt{3},arctanleft(frac{u}{sqrt{3}}right)Bigg|_0^1$$
$$=1-sqrt{3}arctanleft(frac{1}{sqrt{3}}right)=1-frac{sqrt{3}pi}{6}=frac{6-sqrt{3}pi}{6}$$
answered Nov 15 at 21:30
Jevaut
5049
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1
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Hint:
Set $t^2=x,;tge 0$ to obtain the integral of a rational function:
$;mathrm d x=2t,mathrm d t$, $;x=0leftrightarrow t=0$, $;x=1leftrightarrow t=1$, so
$$int_0^1 frac{sqrt{x}}{3+x} ,mathrm d x=int_0^1 frac{2t^2}{t^2+3} ,mathrm d t = 2int_0^1 Bigl(1-frac{3}{t^2+3} Bigr),mathrm d t=dotsm$$
answered Nov 15 at 21:13
Bernard
115k637107
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Make the substitution: $x=u^2, , dx=2u,du$
$$int_0^1 frac{sqrt{x}}{3+x} dx=2int_0^1 frac{u^2}{3+u^2} du={displaystyleint_0^1}left(dfrac{class{steps-node}{cssId{steps-node-4}{u^2+3}}}{u^2+3}-dfrac{class{steps-node}{cssId{steps-node-5}{3}}}{u^2+3}right)mathrm{d}u={displaystyleint_0^1}left(1-dfrac{3}{u^2+3}right)mathrm{d}u$$
$$={displaystyleint_0^1}1,mathrm{d}u-class{steps-node}{cssId{steps-node-6}{3}}{displaystyleint_0^1}dfrac{1}{u^2+3},mathrm{d}u=1-class{steps-node}{cssId{steps-node-7}{dfrac{1}{sqrt{3}}}}{displaystyleint_0^1}dfrac{1}{v^2+1},mathrm{d}v=1-sqrt{3},arctanleft(frac{u}{sqrt{3}}right)Bigg|_0^1$$
$$=1-sqrt{3}arctanleft(frac{1}{sqrt{3}}right)=1-frac{sqrt{3}pi}{6}=frac{6-sqrt{3}pi}{6}$$
answered Nov 15 at 21:30
Jevaut
5049
add a comment |
up vote
1
down vote
accepted
Make the substitution: $x=u^2, , dx=2u,du$
$$int_0^1 frac{sqrt{x}}{3+x} dx=2int_0^1 frac{u^2}{3+u^2} du={displaystyleint_0^1}left(dfrac{class{steps-node}{cssId{steps-node-4}{u^2+3}}}{u^2+3}-dfrac{class{steps-node}{cssId{steps-node-5}{3}}}{u^2+3}right)mathrm{d}u={displaystyleint_0^1}left(1-dfrac{3}{u^2+3}right)mathrm{d}u$$
$$={displaystyleint_0^1}1,mathrm{d}u-class{steps-node}{cssId{steps-node-6}{3}}{displaystyleint_0^1}dfrac{1}{u^2+3},mathrm{d}u=1-class{steps-node}{cssId{steps-node-7}{dfrac{1}{sqrt{3}}}}{displaystyleint_0^1}dfrac{1}{v^2+1},mathrm{d}v=1-sqrt{3},arctanleft(frac{u}{sqrt{3}}right)Bigg|_0^1$$
$$=1-sqrt{3}arctanleft(frac{1}{sqrt{3}}right)=1-frac{sqrt{3}pi}{6}=frac{6-sqrt{3}pi}{6}$$
answered Nov 15 at 21:30
Jevaut
5049
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Make the substitution: $x=u^2, , dx=2u,du$
$$int_0^1 frac{sqrt{x}}{3+x} dx=2int_0^1 frac{u^2}{3+u^2} du={displaystyleint_0^1}left(dfrac{class{steps-node}{cssId{steps-node-4}{u^2+3}}}{u^2+3}-dfrac{class{steps-node}{cssId{steps-node-5}{3}}}{u^2+3}right)mathrm{d}u={displaystyleint_0^1}left(1-dfrac{3}{u^2+3}right)mathrm{d}u$$
$$={displaystyleint_0^1}1,mathrm{d}u-class{steps-node}{cssId{steps-node-6}{3}}{displaystyleint_0^1}dfrac{1}{u^2+3},mathrm{d}u=1-class{steps-node}{cssId{steps-node-7}{dfrac{1}{sqrt{3}}}}{displaystyleint_0^1}dfrac{1}{v^2+1},mathrm{d}v=1-sqrt{3},arctanleft(frac{u}{sqrt{3}}right)Bigg|_0^1$$
$$=1-sqrt{3}arctanleft(frac{1}{sqrt{3}}right)=1-frac{sqrt{3}pi}{6}=frac{6-sqrt{3}pi}{6}$$
answered Nov 15 at 21:30
Jevaut
5049
Make the substitution: $x=u^2, , dx=2u,du$
$$int_0^1 frac{sqrt{x}}{3+x} dx=2int_0^1 frac{u^2}{3+u^2} du={displaystyleint_0^1}left(dfrac{class{steps-node}{cssId{steps-node-4}{u^2+3}}}{u^2+3}-dfrac{class{steps-node}{cssId{steps-node-5}{3}}}{u^2+3}right)mathrm{d}u={displaystyleint_0^1}left(1-dfrac{3}{u^2+3}right)mathrm{d}u$$
$$={displaystyleint_0^1}1,mathrm{d}u-class{steps-node}{cssId{steps-node-6}{3}}{displaystyleint_0^1}dfrac{1}{u^2+3},mathrm{d}u=1-class{steps-node}{cssId{steps-node-7}{dfrac{1}{sqrt{3}}}}{displaystyleint_0^1}dfrac{1}{v^2+1},mathrm{d}v=1-sqrt{3},arctanleft(frac{u}{sqrt{3}}right)Bigg|_0^1$$
$$=1-sqrt{3}arctanleft(frac{1}{sqrt{3}}right)=1-frac{sqrt{3}pi}{6}=frac{6-sqrt{3}pi}{6}$$
answered Nov 15 at 21:30
Jevaut
5049
answered Nov 15 at 21:30
Jevaut
5049
answered Nov 15 at 21:30
Jevaut
5049
answered Nov 15 at 21:30
Jevaut
5049
5049
add a comment |
add a comment |
up vote
1
down vote
Hint:
Set $t^2=x,;tge 0$ to obtain the integral of a rational function:
$;mathrm d x=2t,mathrm d t$, $;x=0leftrightarrow t=0$, $;x=1leftrightarrow t=1$, so
$$int_0^1 frac{sqrt{x}}{3+x} ,mathrm d x=int_0^1 frac{2t^2}{t^2+3} ,mathrm d t = 2int_0^1 Bigl(1-frac{3}{t^2+3} Bigr),mathrm d t=dotsm$$
answered Nov 15 at 21:13
Bernard
115k637107
add a comment |
up vote
1
down vote
Hint:
Set $t^2=x,;tge 0$ to obtain the integral of a rational function:
$;mathrm d x=2t,mathrm d t$, $;x=0leftrightarrow t=0$, $;x=1leftrightarrow t=1$, so
$$int_0^1 frac{sqrt{x}}{3+x} ,mathrm d x=int_0^1 frac{2t^2}{t^2+3} ,mathrm d t = 2int_0^1 Bigl(1-frac{3}{t^2+3} Bigr),mathrm d t=dotsm$$
answered Nov 15 at 21:13
Bernard
115k637107
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint:
Set $t^2=x,;tge 0$ to obtain the integral of a rational function:
$;mathrm d x=2t,mathrm d t$, $;x=0leftrightarrow t=0$, $;x=1leftrightarrow t=1$, so
$$int_0^1 frac{sqrt{x}}{3+x} ,mathrm d x=int_0^1 frac{2t^2}{t^2+3} ,mathrm d t = 2int_0^1 Bigl(1-frac{3}{t^2+3} Bigr),mathrm d t=dotsm$$
answered Nov 15 at 21:13
Bernard
115k637107
Hint:
Set $t^2=x,;tge 0$ to obtain the integral of a rational function:
$;mathrm d x=2t,mathrm d t$, $;x=0leftrightarrow t=0$, $;x=1leftrightarrow t=1$, so
$$int_0^1 frac{sqrt{x}}{3+x} ,mathrm d x=int_0^1 frac{2t^2}{t^2+3} ,mathrm d t = 2int_0^1 Bigl(1-frac{3}{t^2+3} Bigr),mathrm d t=dotsm$$
answered Nov 15 at 21:13
Bernard
115k637107
answered Nov 15 at 21:13
Bernard
115k637107
answered Nov 15 at 21:13
Bernard
115k637107
answered Nov 15 at 21:13
Bernard
115k637107
115k637107
add a comment |
add a comment |
HINT: Change of variables $x=t^2$
– Tito Eliatron
Nov 15 at 20:34
And even, $$x=3t^2$$
– Did
Nov 15 at 20:36