Mixing
limits of definite integral for which definite integral is minimum
$begingroup$
If $displaystyle f(x)=int^{b}_{a}(x^4-2x^2)dx$ is minimum.
Then values of $a$ and $b$ are
Try: I have partial derivative
$$ frac{df}{db}=b^4-2b^2$$
And $$frac{df}{da}=a^4-2a^2$$
For maximum and minimum $displaystyle frac{df}{da}=0$ and $displaystyle frac{df}{db}=0$
So $a^4-2a^2=0Rightarrow a=0,pm sqrt{2}.$
and $b^4-2b^2=0Rightarrow b=0,pmsqrt{2}.$
So we have $(a,b)=bigg{(0,0),bigg(-sqrt{2},0bigg),bigg(0,sqrt{2}bigg)bigg}$
But answer given as $(a,b)=bigg(-sqrt{2},sqrt{2}bigg)$
Could some help me How interval is $bigg(-sqrt{2},sqrt{2}bigg)$
thanks in advance
calculus
asked Jan 11 at 8:50
DXTDXT
5,6842630
$endgroup$
add a comment |
$begingroup$
If $displaystyle f(x)=int^{b}_{a}(x^4-2x^2)dx$ is minimum.
Then values of $a$ and $b$ are
Try: I have partial derivative
$$ frac{df}{db}=b^4-2b^2$$
And $$frac{df}{da}=a^4-2a^2$$
For maximum and minimum $displaystyle frac{df}{da}=0$ and $displaystyle frac{df}{db}=0$
So $a^4-2a^2=0Rightarrow a=0,pm sqrt{2}.$
and $b^4-2b^2=0Rightarrow b=0,pmsqrt{2}.$
So we have $(a,b)=bigg{(0,0),bigg(-sqrt{2},0bigg),bigg(0,sqrt{2}bigg)bigg}$
But answer given as $(a,b)=bigg(-sqrt{2},sqrt{2}bigg)$
Could some help me How interval is $bigg(-sqrt{2},sqrt{2}bigg)$
thanks in advance
calculus
asked Jan 11 at 8:50
DXTDXT
5,6842630
$endgroup$
$begingroup$
You can also proceed by showing that $x^4-2x^2leq 0$ only on that interval.
$endgroup$
– projectilemotion
Jan 11 at 8:57
$begingroup$
$frac{df}{da}=2a^2-a^4$
$endgroup$
– Aleksas Domarkas
Jan 11 at 9:22
add a comment |
$begingroup$
If $displaystyle f(x)=int^{b}_{a}(x^4-2x^2)dx$ is minimum.
Then values of $a$ and $b$ are
Try: I have partial derivative
$$ frac{df}{db}=b^4-2b^2$$
And $$frac{df}{da}=a^4-2a^2$$
For maximum and minimum $displaystyle frac{df}{da}=0$ and $displaystyle frac{df}{db}=0$
So $a^4-2a^2=0Rightarrow a=0,pm sqrt{2}.$
and $b^4-2b^2=0Rightarrow b=0,pmsqrt{2}.$
So we have $(a,b)=bigg{(0,0),bigg(-sqrt{2},0bigg),bigg(0,sqrt{2}bigg)bigg}$
But answer given as $(a,b)=bigg(-sqrt{2},sqrt{2}bigg)$
Could some help me How interval is $bigg(-sqrt{2},sqrt{2}bigg)$
thanks in advance
calculus
asked Jan 11 at 8:50
DXTDXT
5,6842630
$endgroup$
If $displaystyle f(x)=int^{b}_{a}(x^4-2x^2)dx$ is minimum.
Then values of $a$ and $b$ are
Try: I have partial derivative
$$ frac{df}{db}=b^4-2b^2$$
And $$frac{df}{da}=a^4-2a^2$$
For maximum and minimum $displaystyle frac{df}{da}=0$ and $displaystyle frac{df}{db}=0$
So $a^4-2a^2=0Rightarrow a=0,pm sqrt{2}.$
and $b^4-2b^2=0Rightarrow b=0,pmsqrt{2}.$
So we have $(a,b)=bigg{(0,0),bigg(-sqrt{2},0bigg),bigg(0,sqrt{2}bigg)bigg}$
But answer given as $(a,b)=bigg(-sqrt{2},sqrt{2}bigg)$
Could some help me How interval is $bigg(-sqrt{2},sqrt{2}bigg)$
thanks in advance
calculus
calculus
asked Jan 11 at 8:50
DXTDXT
5,6842630
asked Jan 11 at 8:50
DXTDXT
5,6842630
asked Jan 11 at 8:50
DXTDXT
5,6842630
asked Jan 11 at 8:50
DXTDXT
5,6842630
asked Jan 11 at 8:50
DXTDXT
5,6842630
5,6842630
$begingroup$
You can also proceed by showing that $x^4-2x^2leq 0$ only on that interval.
$endgroup$
– projectilemotion
Jan 11 at 8:57
$begingroup$
$frac{df}{da}=2a^2-a^4$
$endgroup$
– Aleksas Domarkas
Jan 11 at 9:22
add a comment |
$begingroup$
You can also proceed by showing that $x^4-2x^2leq 0$ only on that interval.
$endgroup$
– projectilemotion
Jan 11 at 8:57
$begingroup$
$frac{df}{da}=2a^2-a^4$
$endgroup$
– Aleksas Domarkas
Jan 11 at 9:22
$begingroup$
You can also proceed by showing that $x^4-2x^2leq 0$ only on that interval.
$endgroup$
– projectilemotion
Jan 11 at 8:57
$begingroup$
You can also proceed by showing that $x^4-2x^2leq 0$ only on that interval.
$endgroup$
– projectilemotion
Jan 11 at 8:57
$begingroup$
$frac{df}{da}=2a^2-a^4$
$endgroup$
– Aleksas Domarkas
Jan 11 at 9:22
$begingroup$
$frac{df}{da}=2a^2-a^4$
$endgroup$
– Aleksas Domarkas
Jan 11 at 9:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Like any optimization problem, the critical points give you a list of possibilities. (Actually, endpoints of $pminfty$ should be in there as well - they're the supremum, as those integrals diverge to $infty$.)
It's still only a list of possibilities. Among the intervals with both endpoints in ${-sqrt{2},0,sqrt{2}}$ - six possibilities - we still have to find which one produces the smallest (most negative) integral. As it turns out, it's $(-sqrt{2},sqrt{2})$, the largest of those possible intervals.
Oh, and we do need the assumption that $ale b$ here. Without it, $int_{infty}^{-infty} x^4-2x^2,dx=-infty$ overwhelms what we're looking for.
answered Jan 11 at 9:02
jmerryjmerry
7,593921
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
$begingroup$
Like any optimization problem, the critical points give you a list of possibilities. (Actually, endpoints of $pminfty$ should be in there as well - they're the supremum, as those integrals diverge to $infty$.)
It's still only a list of possibilities. Among the intervals with both endpoints in ${-sqrt{2},0,sqrt{2}}$ - six possibilities - we still have to find which one produces the smallest (most negative) integral. As it turns out, it's $(-sqrt{2},sqrt{2})$, the largest of those possible intervals.
Oh, and we do need the assumption that $ale b$ here. Without it, $int_{infty}^{-infty} x^4-2x^2,dx=-infty$ overwhelms what we're looking for.
answered Jan 11 at 9:02
jmerryjmerry
7,593921
$endgroup$
add a comment |
$begingroup$
Like any optimization problem, the critical points give you a list of possibilities. (Actually, endpoints of $pminfty$ should be in there as well - they're the supremum, as those integrals diverge to $infty$.)
It's still only a list of possibilities. Among the intervals with both endpoints in ${-sqrt{2},0,sqrt{2}}$ - six possibilities - we still have to find which one produces the smallest (most negative) integral. As it turns out, it's $(-sqrt{2},sqrt{2})$, the largest of those possible intervals.
Oh, and we do need the assumption that $ale b$ here. Without it, $int_{infty}^{-infty} x^4-2x^2,dx=-infty$ overwhelms what we're looking for.
answered Jan 11 at 9:02
jmerryjmerry
7,593921
$endgroup$
add a comment |
$begingroup$
Like any optimization problem, the critical points give you a list of possibilities. (Actually, endpoints of $pminfty$ should be in there as well - they're the supremum, as those integrals diverge to $infty$.)
It's still only a list of possibilities. Among the intervals with both endpoints in ${-sqrt{2},0,sqrt{2}}$ - six possibilities - we still have to find which one produces the smallest (most negative) integral. As it turns out, it's $(-sqrt{2},sqrt{2})$, the largest of those possible intervals.
Oh, and we do need the assumption that $ale b$ here. Without it, $int_{infty}^{-infty} x^4-2x^2,dx=-infty$ overwhelms what we're looking for.
answered Jan 11 at 9:02
jmerryjmerry
7,593921
$endgroup$
Like any optimization problem, the critical points give you a list of possibilities. (Actually, endpoints of $pminfty$ should be in there as well - they're the supremum, as those integrals diverge to $infty$.)
It's still only a list of possibilities. Among the intervals with both endpoints in ${-sqrt{2},0,sqrt{2}}$ - six possibilities - we still have to find which one produces the smallest (most negative) integral. As it turns out, it's $(-sqrt{2},sqrt{2})$, the largest of those possible intervals.
Oh, and we do need the assumption that $ale b$ here. Without it, $int_{infty}^{-infty} x^4-2x^2,dx=-infty$ overwhelms what we're looking for.
answered Jan 11 at 9:02
jmerryjmerry
7,593921
answered Jan 11 at 9:02
jmerryjmerry
7,593921
answered Jan 11 at 9:02
jmerryjmerry
7,593921
answered Jan 11 at 9:02
jmerryjmerry
7,593921
7,593921
add a comment |
add a comment |
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$begingroup$
You can also proceed by showing that $x^4-2x^2leq 0$ only on that interval.
$endgroup$
– projectilemotion
Jan 11 at 8:57
$begingroup$
$frac{df}{da}=2a^2-a^4$
$endgroup$
– Aleksas Domarkas
Jan 11 at 9:22