using Leibniz rule to solve definite integralLeibniz integral rule implementationDerivative w.r.t. limits of integration of double definite integral (Leibniz' rule)Leibniz rule for improper integralA general case - Leibniz ruleMisconception in applying Leibniz' ruleDerivative of Parametric Integral Using LeibnizApplying Leibniz rule to multiple integralUsing Leibniz Rule to Find an ExpressionLeibniz integral rule in SDEWhy can I not use leibniz integral rule in this case?
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using Leibniz rule to solve definite integral
Leibniz integral rule implementationDerivative w.r.t. limits of integration of double definite integral (Leibniz' rule)Leibniz rule for improper integralA general case - Leibniz ruleMisconception in applying Leibniz' ruleDerivative of Parametric Integral Using LeibnizApplying Leibniz rule to multiple integralUsing Leibniz Rule to Find an ExpressionLeibniz integral rule in SDEWhy can I not use leibniz integral rule in this case?
$begingroup$
The total number of distinct $x$ in $[0,1]$ for which $$int_0^xfract^21+t^4dt=2x-1$$ holds is
I tried solving using Leibniz rule but that gives $0$ no. of solutions. However the answer is $1$.
integration definite-integrals
edited 9 hours ago
cmk
62010
asked 9 hours ago
LakshayLakshay
1306
$endgroup$
add a comment |
$begingroup$
The total number of distinct $x$ in $[0,1]$ for which $$int_0^xfract^21+t^4dt=2x-1$$ holds is
I tried solving using Leibniz rule but that gives $0$ no. of solutions. However the answer is $1$.
integration definite-integrals
edited 9 hours ago
cmk
62010
asked 9 hours ago
LakshayLakshay
1306
$endgroup$
$begingroup$
Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
$endgroup$
– cmk
9 hours ago
$begingroup$
why can't i use differentiation
$endgroup$
– Lakshay
9 hours ago
1
$begingroup$
You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
$endgroup$
– Vizag
9 hours ago
$begingroup$
This question was asked in IIT JEE (A) in 2016.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
add a comment |
$begingroup$
The total number of distinct $x$ in $[0,1]$ for which $$int_0^xfract^21+t^4dt=2x-1$$ holds is
I tried solving using Leibniz rule but that gives $0$ no. of solutions. However the answer is $1$.
integration definite-integrals
edited 9 hours ago
cmk
62010
asked 9 hours ago
LakshayLakshay
1306
$endgroup$
The total number of distinct $x$ in $[0,1]$ for which $$int_0^xfract^21+t^4dt=2x-1$$ holds is
I tried solving using Leibniz rule but that gives $0$ no. of solutions. However the answer is $1$.
integration definite-integrals
integration definite-integrals
edited 9 hours ago
cmk
62010
asked 9 hours ago
LakshayLakshay
1306
edited 9 hours ago
cmk
62010
asked 9 hours ago
LakshayLakshay
1306
edited 9 hours ago
cmk
62010
edited 9 hours ago
cmk
62010
edited 9 hours ago
cmk
62010
62010
asked 9 hours ago
LakshayLakshay
1306
asked 9 hours ago
LakshayLakshay
1306
asked 9 hours ago
LakshayLakshay
1306
1306
$begingroup$
Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
$endgroup$
– cmk
9 hours ago
$begingroup$
why can't i use differentiation
$endgroup$
– Lakshay
9 hours ago
1
$begingroup$
You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
$endgroup$
– Vizag
9 hours ago
$begingroup$
This question was asked in IIT JEE (A) in 2016.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
add a comment |
$begingroup$
Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
$endgroup$
– cmk
9 hours ago
$begingroup$
why can't i use differentiation
$endgroup$
– Lakshay
9 hours ago
1
$begingroup$
You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
$endgroup$
– Vizag
9 hours ago
$begingroup$
This question was asked in IIT JEE (A) in 2016.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
$begingroup$
Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
$endgroup$
– cmk
9 hours ago
$begingroup$
Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
$endgroup$
– cmk
9 hours ago
$begingroup$
why can't i use differentiation
$endgroup$
– Lakshay
9 hours ago
$begingroup$
why can't i use differentiation
$endgroup$
– Lakshay
9 hours ago
1
1
$begingroup$
You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
$endgroup$
– Vizag
9 hours ago
$begingroup$
You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
$endgroup$
– Vizag
9 hours ago
$begingroup$
This question was asked in IIT JEE (A) in 2016.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
$begingroup$
This question was asked in IIT JEE (A) in 2016.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the function $$f(x)=2x-1-intlimits_0^xfract^21+t^4 dt.$$ Clearly, $f$ is continuous on $[0,1].$ Note that $f(0)=-1<0$, and $$f(1)=1-intlimits_0^1fract^21+t^4 dtgeq 1-intlimits_0^1frac11+0 dt=0.$$ If $f(1)$ is zero, we've found a solution. If not, then $f(1)>0.$ In this case, the intermediate value theorem guarantees that there exists $cin (0,1)$ for which $f(c)=0,$ or $$2c-1=intlimits_0^cfract^21+t^4 dt.$$ This shows that there is, at least, one solution. Next, note that $$f'(x)=2-fracx^21+x^4>0$$ for $xin [0,1]$, so the function is strictly increasing, which tells us that the previously-found zero of $f$ is the only such zero. In the end, we did use the fundamental theorem of calculus here, but only to show that the zero of $f$ was unique.
answered 9 hours ago
cmkcmk
62010
$endgroup$
add a comment |
$begingroup$
Hint: Use that $$1+t^4=1+t^4+2t^2-2t^2=(1+t^2)^2-(sqrt2t)^2=(1+t^2-sqrt2t)(1+t^2+sqrt2t)$$
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.3k42867
$endgroup$
$begingroup$
it complicates the solution.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the function $$f(x)=2x-1-intlimits_0^xfract^21+t^4 dt.$$ Clearly, $f$ is continuous on $[0,1].$ Note that $f(0)=-1<0$, and $$f(1)=1-intlimits_0^1fract^21+t^4 dtgeq 1-intlimits_0^1frac11+0 dt=0.$$ If $f(1)$ is zero, we've found a solution. If not, then $f(1)>0.$ In this case, the intermediate value theorem guarantees that there exists $cin (0,1)$ for which $f(c)=0,$ or $$2c-1=intlimits_0^cfract^21+t^4 dt.$$ This shows that there is, at least, one solution. Next, note that $$f'(x)=2-fracx^21+x^4>0$$ for $xin [0,1]$, so the function is strictly increasing, which tells us that the previously-found zero of $f$ is the only such zero. In the end, we did use the fundamental theorem of calculus here, but only to show that the zero of $f$ was unique.
answered 9 hours ago
cmkcmk
62010
$endgroup$
add a comment |
$begingroup$
Consider the function $$f(x)=2x-1-intlimits_0^xfract^21+t^4 dt.$$ Clearly, $f$ is continuous on $[0,1].$ Note that $f(0)=-1<0$, and $$f(1)=1-intlimits_0^1fract^21+t^4 dtgeq 1-intlimits_0^1frac11+0 dt=0.$$ If $f(1)$ is zero, we've found a solution. If not, then $f(1)>0.$ In this case, the intermediate value theorem guarantees that there exists $cin (0,1)$ for which $f(c)=0,$ or $$2c-1=intlimits_0^cfract^21+t^4 dt.$$ This shows that there is, at least, one solution. Next, note that $$f'(x)=2-fracx^21+x^4>0$$ for $xin [0,1]$, so the function is strictly increasing, which tells us that the previously-found zero of $f$ is the only such zero. In the end, we did use the fundamental theorem of calculus here, but only to show that the zero of $f$ was unique.
answered 9 hours ago
cmkcmk
62010
$endgroup$
add a comment |
$begingroup$
Consider the function $$f(x)=2x-1-intlimits_0^xfract^21+t^4 dt.$$ Clearly, $f$ is continuous on $[0,1].$ Note that $f(0)=-1<0$, and $$f(1)=1-intlimits_0^1fract^21+t^4 dtgeq 1-intlimits_0^1frac11+0 dt=0.$$ If $f(1)$ is zero, we've found a solution. If not, then $f(1)>0.$ In this case, the intermediate value theorem guarantees that there exists $cin (0,1)$ for which $f(c)=0,$ or $$2c-1=intlimits_0^cfract^21+t^4 dt.$$ This shows that there is, at least, one solution. Next, note that $$f'(x)=2-fracx^21+x^4>0$$ for $xin [0,1]$, so the function is strictly increasing, which tells us that the previously-found zero of $f$ is the only such zero. In the end, we did use the fundamental theorem of calculus here, but only to show that the zero of $f$ was unique.
answered 9 hours ago
cmkcmk
62010
$endgroup$
Consider the function $$f(x)=2x-1-intlimits_0^xfract^21+t^4 dt.$$ Clearly, $f$ is continuous on $[0,1].$ Note that $f(0)=-1<0$, and $$f(1)=1-intlimits_0^1fract^21+t^4 dtgeq 1-intlimits_0^1frac11+0 dt=0.$$ If $f(1)$ is zero, we've found a solution. If not, then $f(1)>0.$ In this case, the intermediate value theorem guarantees that there exists $cin (0,1)$ for which $f(c)=0,$ or $$2c-1=intlimits_0^cfract^21+t^4 dt.$$ This shows that there is, at least, one solution. Next, note that $$f'(x)=2-fracx^21+x^4>0$$ for $xin [0,1]$, so the function is strictly increasing, which tells us that the previously-found zero of $f$ is the only such zero. In the end, we did use the fundamental theorem of calculus here, but only to show that the zero of $f$ was unique.
answered 9 hours ago
cmkcmk
62010
edited 8 hours ago
answered 9 hours ago
cmkcmk
62010
answered 9 hours ago
cmkcmk
62010
answered 9 hours ago
cmkcmk
62010
62010
add a comment |
add a comment |
$begingroup$
Hint: Use that $$1+t^4=1+t^4+2t^2-2t^2=(1+t^2)^2-(sqrt2t)^2=(1+t^2-sqrt2t)(1+t^2+sqrt2t)$$
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.3k42867
$endgroup$
$begingroup$
it complicates the solution.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
add a comment |
$begingroup$
Hint: Use that $$1+t^4=1+t^4+2t^2-2t^2=(1+t^2)^2-(sqrt2t)^2=(1+t^2-sqrt2t)(1+t^2+sqrt2t)$$
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.3k42867
$endgroup$
$begingroup$
it complicates the solution.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
add a comment |
$begingroup$
Hint: Use that $$1+t^4=1+t^4+2t^2-2t^2=(1+t^2)^2-(sqrt2t)^2=(1+t^2-sqrt2t)(1+t^2+sqrt2t)$$
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.3k42867
$endgroup$
Hint: Use that $$1+t^4=1+t^4+2t^2-2t^2=(1+t^2)^2-(sqrt2t)^2=(1+t^2-sqrt2t)(1+t^2+sqrt2t)$$
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.3k42867
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.3k42867
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.3k42867
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.3k42867
81.3k42867
$begingroup$
it complicates the solution.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
add a comment |
$begingroup$
it complicates the solution.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
$begingroup$
it complicates the solution.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
$begingroup$
it complicates the solution.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago
add a comment |
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$begingroup$
Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
$endgroup$
– cmk
9 hours ago
$begingroup$
why can't i use differentiation
$endgroup$
– Lakshay
9 hours ago
1
$begingroup$
You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
$endgroup$
– Vizag
9 hours ago
$begingroup$
This question was asked in IIT JEE (A) in 2016.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago