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Evaluate definite integral using limit of summations
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I need to evaluate a definite integral using sums.
$$
int_-5^5 (x-sqrt25-x^2)dx
$$
I start by getting de value of $Delta x$:
$$
Delta x = frac5-(-5)n=frac10n
$$
Then for $x^*_i$:
$$
x^*_i=-5+Delta xi=frac10in-5
$$
Then I proceed to $f(x^*_i)$:
$$
f(x^*_i)=f(frac10in-5) \
=frac10in-5-sqrt25-(frac10in-5)^2 \
=frac10i-5n-10sqrti(n-i)n
$$
And then, finally, the sum:
$$
lim_x to inftysum_i=1^nfrac10i-5n-10sqrti(n-i)nfrac10n \
=lim_x to inftyfrac100n(sum_i=1^ni-fracn2sum_i=1^n1-sum_i=1^nsqrti(n-i))
$$
If my algebra was correct and I didn't blunder somewhere, that is. But I'm stuck there, I mean I can use summation identities for the first sum and the second one, but I don't have anything for the summation of square root.
How can I move on from here, or am I taking the wrong approach? Thanks in advance.
calculus integration definite-integrals
asked Jul 28 at 6:01
Fernando Gómez
947
add a comment |Â
up vote
3
down vote
favorite
I need to evaluate a definite integral using sums.
$$
int_-5^5 (x-sqrt25-x^2)dx
$$
I start by getting de value of $Delta x$:
$$
Delta x = frac5-(-5)n=frac10n
$$
Then for $x^*_i$:
$$
x^*_i=-5+Delta xi=frac10in-5
$$
Then I proceed to $f(x^*_i)$:
$$
f(x^*_i)=f(frac10in-5) \
=frac10in-5-sqrt25-(frac10in-5)^2 \
=frac10i-5n-10sqrti(n-i)n
$$
And then, finally, the sum:
$$
lim_x to inftysum_i=1^nfrac10i-5n-10sqrti(n-i)nfrac10n \
=lim_x to inftyfrac100n(sum_i=1^ni-fracn2sum_i=1^n1-sum_i=1^nsqrti(n-i))
$$
If my algebra was correct and I didn't blunder somewhere, that is. But I'm stuck there, I mean I can use summation identities for the first sum and the second one, but I don't have anything for the summation of square root.
How can I move on from here, or am I taking the wrong approach? Thanks in advance.
calculus integration definite-integrals
asked Jul 28 at 6:01
Fernando Gómez
947
1
Well, you can use the linearity of the integral and split it into $int x dx$ and $intsqrt25-x^2dx$ and then the second part is even, so you can simplify a bit, doesn't that help?
– Michal Dvořák
Jul 28 at 7:25
For the sum of square roots, see Sum of Square roots formula.
– user539887
Jul 28 at 7:51
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I need to evaluate a definite integral using sums.
$$
int_-5^5 (x-sqrt25-x^2)dx
$$
I start by getting de value of $Delta x$:
$$
Delta x = frac5-(-5)n=frac10n
$$
Then for $x^*_i$:
$$
x^*_i=-5+Delta xi=frac10in-5
$$
Then I proceed to $f(x^*_i)$:
$$
f(x^*_i)=f(frac10in-5) \
=frac10in-5-sqrt25-(frac10in-5)^2 \
=frac10i-5n-10sqrti(n-i)n
$$
And then, finally, the sum:
$$
lim_x to inftysum_i=1^nfrac10i-5n-10sqrti(n-i)nfrac10n \
=lim_x to inftyfrac100n(sum_i=1^ni-fracn2sum_i=1^n1-sum_i=1^nsqrti(n-i))
$$
If my algebra was correct and I didn't blunder somewhere, that is. But I'm stuck there, I mean I can use summation identities for the first sum and the second one, but I don't have anything for the summation of square root.
How can I move on from here, or am I taking the wrong approach? Thanks in advance.
calculus integration definite-integrals
asked Jul 28 at 6:01
Fernando Gómez
947
I need to evaluate a definite integral using sums.
$$
int_-5^5 (x-sqrt25-x^2)dx
$$
I start by getting de value of $Delta x$:
$$
Delta x = frac5-(-5)n=frac10n
$$
Then for $x^*_i$:
$$
x^*_i=-5+Delta xi=frac10in-5
$$
Then I proceed to $f(x^*_i)$:
$$
f(x^*_i)=f(frac10in-5) \
=frac10in-5-sqrt25-(frac10in-5)^2 \
=frac10i-5n-10sqrti(n-i)n
$$
And then, finally, the sum:
$$
lim_x to inftysum_i=1^nfrac10i-5n-10sqrti(n-i)nfrac10n \
=lim_x to inftyfrac100n(sum_i=1^ni-fracn2sum_i=1^n1-sum_i=1^nsqrti(n-i))
$$
If my algebra was correct and I didn't blunder somewhere, that is. But I'm stuck there, I mean I can use summation identities for the first sum and the second one, but I don't have anything for the summation of square root.
How can I move on from here, or am I taking the wrong approach? Thanks in advance.
calculus integration definite-integrals
asked Jul 28 at 6:01
Fernando Gómez
947
asked Jul 28 at 6:01
Fernando Gómez
947
asked Jul 28 at 6:01
Fernando Gómez
947
asked Jul 28 at 6:01
Fernando Gómez
947
947
1
Well, you can use the linearity of the integral and split it into $int x dx$ and $intsqrt25-x^2dx$ and then the second part is even, so you can simplify a bit, doesn't that help?
– Michal Dvořák
Jul 28 at 7:25
For the sum of square roots, see Sum of Square roots formula.
– user539887
Jul 28 at 7:51
add a comment |Â
1
Well, you can use the linearity of the integral and split it into $int x dx$ and $intsqrt25-x^2dx$ and then the second part is even, so you can simplify a bit, doesn't that help?
– Michal Dvořák
Jul 28 at 7:25
For the sum of square roots, see Sum of Square roots formula.
– user539887
Jul 28 at 7:51
1
1
Well, you can use the linearity of the integral and split it into $int x dx$ and $intsqrt25-x^2dx$ and then the second part is even, so you can simplify a bit, doesn't that help?
– Michal Dvořák
Jul 28 at 7:25
Well, you can use the linearity of the integral and split it into $int x dx$ and $intsqrt25-x^2dx$ and then the second part is even, so you can simplify a bit, doesn't that help?
– Michal Dvořák
Jul 28 at 7:25
For the sum of square roots, see Sum of Square roots formula.
– user539887
Jul 28 at 7:51
For the sum of square roots, see Sum of Square roots formula.
– user539887
Jul 28 at 7:51
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
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accepted
You can calcuate these two sums independently:
$$int_-5^5 (x-sqrt25-x^2)dx=int_-5^5 xdx-int_-5^5 sqrt25-x^2dx$$
Let's calculate the first integral. Obviously it's zero because the function $f(x)=x$ is odd but if you insist you can prove it by summation:
Divide interval from -5 to +5 in $n$ equal segments:
$$Delta x_i=frac10n, quad y_i=x_i=-5+frac 10in,quad i=0, 1, 2,...,n $$
$$A_n=sum_i=0^n-1y_iDelta x_i=sum_i=0^n-1(-5+frac 10in)frac10n$$
$$=-50sum_i=0^n-1frac1n+frac100n^2sum_i=0^n-1i=-50+frac100n^2fracn(n-1)2=-50+50fracn-1n=-frac50n$$
So the first integral is:
$$I_1=lim_nto inftyA_n=lim_nto inftyfrac-50n=0$$
Let's tackle the second one:
$$I_2=int_-5^5 sqrt25-x^2dx$$
The graph of function $f(x)=sqrt25-x^2$ is symmetric with respect to $y$-axis so:
$$I_2=2int_0^5 sqrt25-x^2dx$$
Let's calculate:
$$I_3=int_0^5 sqrt25-x^2dx$$
...by symmation. Again, I will divide the interval of integration into $n$ segments but this time they won't be of equal length. I will introduce variable $varphi$ such that:
$$Deltavarphi=fracpi2n, quad varphi_i=iDeltavarphi=fracipi2n, quad i=0,1,2,...,n$$
$$x_i=5sinvarphi_i$$
$$y_i=sqrt25-x_i^2=5cosvarphi_i$$
$$Delta x_i=x_i+1-x_i=5(sinvarphi_i+1-sinvarphi_i)=5(sin(varphi_i+Deltavarphi)-sinvarphi_i)=5(sinvarphi_i(cosDeltavarphi-1)+cosvarphi_isinDeltavarphi)=5(-2sin^2fracDeltavarphi2sinvarphi_i+cosvarphi_isinDeltavarphi)$$
Having in mind that for small values of $Deltavarphi$:
$$sin Deltavarphi=Deltavarphi+O(Deltavarphi^3)$$
...we can write:
$$Delta x_i=5cosvarphi_iDeltavarphi$$
Now the integral becomes:
$$Delta A_i=y_iDelta x_i=25Deltavarphicos^2varphi_i=frac25Deltavarphi2(1+cos2varphi_i)=frac25Deltavarphi2(1+cosfracipin)$$
$$A_n=sum_i=0^n Delta A_i=frac25Deltavarphi2sum_i=0^n-11+frac25Deltavarphi2sum_i=0^n-1 cosfracipin=frac25nDeltavarphi2+frac25Deltavarphi2sum_i=0^n-1 cosfracipin$$
$$=frac25pi4+frac25pi4nsum_i=0^n-1 cosfracipin$$
By using trigonometric identity of Lagrange you can show that:
$$sum_i=0^n-1 cosfracipin=1+sum_i=1^n-1 cosfracipin=1-frac12+fracsin((n-1+frac12)fracpin)2sinfracpi2n=frac12+fracsin(pi-fracpi2n)2sinfracpi2n=1$$
It means that:
$$A_n=frac25pi4+frac25pi4n$$
$$I_3=lim_ntoinftyA_n=lim_ntoinftyfrac25pi4+frac25pi4n=frac25pi4$$
$$I_2=2I_3=frac25pi2$$
answered Jul 28 at 14:28
Oldboy
2,6001316
Thank you that was very thorough. I had to read it several times but I think I got it. Thanks again!
– Fernando Gómez
Jul 28 at 22:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You can calcuate these two sums independently:
$$int_-5^5 (x-sqrt25-x^2)dx=int_-5^5 xdx-int_-5^5 sqrt25-x^2dx$$
Let's calculate the first integral. Obviously it's zero because the function $f(x)=x$ is odd but if you insist you can prove it by summation:
Divide interval from -5 to +5 in $n$ equal segments:
$$Delta x_i=frac10n, quad y_i=x_i=-5+frac 10in,quad i=0, 1, 2,...,n $$
$$A_n=sum_i=0^n-1y_iDelta x_i=sum_i=0^n-1(-5+frac 10in)frac10n$$
$$=-50sum_i=0^n-1frac1n+frac100n^2sum_i=0^n-1i=-50+frac100n^2fracn(n-1)2=-50+50fracn-1n=-frac50n$$
So the first integral is:
$$I_1=lim_nto inftyA_n=lim_nto inftyfrac-50n=0$$
Let's tackle the second one:
$$I_2=int_-5^5 sqrt25-x^2dx$$
The graph of function $f(x)=sqrt25-x^2$ is symmetric with respect to $y$-axis so:
$$I_2=2int_0^5 sqrt25-x^2dx$$
Let's calculate:
$$I_3=int_0^5 sqrt25-x^2dx$$
...by symmation. Again, I will divide the interval of integration into $n$ segments but this time they won't be of equal length. I will introduce variable $varphi$ such that:
$$Deltavarphi=fracpi2n, quad varphi_i=iDeltavarphi=fracipi2n, quad i=0,1,2,...,n$$
$$x_i=5sinvarphi_i$$
$$y_i=sqrt25-x_i^2=5cosvarphi_i$$
$$Delta x_i=x_i+1-x_i=5(sinvarphi_i+1-sinvarphi_i)=5(sin(varphi_i+Deltavarphi)-sinvarphi_i)=5(sinvarphi_i(cosDeltavarphi-1)+cosvarphi_isinDeltavarphi)=5(-2sin^2fracDeltavarphi2sinvarphi_i+cosvarphi_isinDeltavarphi)$$
Having in mind that for small values of $Deltavarphi$:
$$sin Deltavarphi=Deltavarphi+O(Deltavarphi^3)$$
...we can write:
$$Delta x_i=5cosvarphi_iDeltavarphi$$
Now the integral becomes:
$$Delta A_i=y_iDelta x_i=25Deltavarphicos^2varphi_i=frac25Deltavarphi2(1+cos2varphi_i)=frac25Deltavarphi2(1+cosfracipin)$$
$$A_n=sum_i=0^n Delta A_i=frac25Deltavarphi2sum_i=0^n-11+frac25Deltavarphi2sum_i=0^n-1 cosfracipin=frac25nDeltavarphi2+frac25Deltavarphi2sum_i=0^n-1 cosfracipin$$
$$=frac25pi4+frac25pi4nsum_i=0^n-1 cosfracipin$$
By using trigonometric identity of Lagrange you can show that:
$$sum_i=0^n-1 cosfracipin=1+sum_i=1^n-1 cosfracipin=1-frac12+fracsin((n-1+frac12)fracpin)2sinfracpi2n=frac12+fracsin(pi-fracpi2n)2sinfracpi2n=1$$
It means that:
$$A_n=frac25pi4+frac25pi4n$$
$$I_3=lim_ntoinftyA_n=lim_ntoinftyfrac25pi4+frac25pi4n=frac25pi4$$
$$I_2=2I_3=frac25pi2$$
answered Jul 28 at 14:28
Oldboy
2,6001316
Thank you that was very thorough. I had to read it several times but I think I got it. Thanks again!
– Fernando Gómez
Jul 28 at 22:31
add a comment |Â
up vote
2
down vote
accepted
You can calcuate these two sums independently:
$$int_-5^5 (x-sqrt25-x^2)dx=int_-5^5 xdx-int_-5^5 sqrt25-x^2dx$$
Let's calculate the first integral. Obviously it's zero because the function $f(x)=x$ is odd but if you insist you can prove it by summation:
Divide interval from -5 to +5 in $n$ equal segments:
$$Delta x_i=frac10n, quad y_i=x_i=-5+frac 10in,quad i=0, 1, 2,...,n $$
$$A_n=sum_i=0^n-1y_iDelta x_i=sum_i=0^n-1(-5+frac 10in)frac10n$$
$$=-50sum_i=0^n-1frac1n+frac100n^2sum_i=0^n-1i=-50+frac100n^2fracn(n-1)2=-50+50fracn-1n=-frac50n$$
So the first integral is:
$$I_1=lim_nto inftyA_n=lim_nto inftyfrac-50n=0$$
Let's tackle the second one:
$$I_2=int_-5^5 sqrt25-x^2dx$$
The graph of function $f(x)=sqrt25-x^2$ is symmetric with respect to $y$-axis so:
$$I_2=2int_0^5 sqrt25-x^2dx$$
Let's calculate:
$$I_3=int_0^5 sqrt25-x^2dx$$
...by symmation. Again, I will divide the interval of integration into $n$ segments but this time they won't be of equal length. I will introduce variable $varphi$ such that:
$$Deltavarphi=fracpi2n, quad varphi_i=iDeltavarphi=fracipi2n, quad i=0,1,2,...,n$$
$$x_i=5sinvarphi_i$$
$$y_i=sqrt25-x_i^2=5cosvarphi_i$$
$$Delta x_i=x_i+1-x_i=5(sinvarphi_i+1-sinvarphi_i)=5(sin(varphi_i+Deltavarphi)-sinvarphi_i)=5(sinvarphi_i(cosDeltavarphi-1)+cosvarphi_isinDeltavarphi)=5(-2sin^2fracDeltavarphi2sinvarphi_i+cosvarphi_isinDeltavarphi)$$
Having in mind that for small values of $Deltavarphi$:
$$sin Deltavarphi=Deltavarphi+O(Deltavarphi^3)$$
...we can write:
$$Delta x_i=5cosvarphi_iDeltavarphi$$
Now the integral becomes:
$$Delta A_i=y_iDelta x_i=25Deltavarphicos^2varphi_i=frac25Deltavarphi2(1+cos2varphi_i)=frac25Deltavarphi2(1+cosfracipin)$$
$$A_n=sum_i=0^n Delta A_i=frac25Deltavarphi2sum_i=0^n-11+frac25Deltavarphi2sum_i=0^n-1 cosfracipin=frac25nDeltavarphi2+frac25Deltavarphi2sum_i=0^n-1 cosfracipin$$
$$=frac25pi4+frac25pi4nsum_i=0^n-1 cosfracipin$$
By using trigonometric identity of Lagrange you can show that:
$$sum_i=0^n-1 cosfracipin=1+sum_i=1^n-1 cosfracipin=1-frac12+fracsin((n-1+frac12)fracpin)2sinfracpi2n=frac12+fracsin(pi-fracpi2n)2sinfracpi2n=1$$
It means that:
$$A_n=frac25pi4+frac25pi4n$$
$$I_3=lim_ntoinftyA_n=lim_ntoinftyfrac25pi4+frac25pi4n=frac25pi4$$
$$I_2=2I_3=frac25pi2$$
answered Jul 28 at 14:28
Oldboy
2,6001316
Thank you that was very thorough. I had to read it several times but I think I got it. Thanks again!
– Fernando Gómez
Jul 28 at 22:31
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You can calcuate these two sums independently:
$$int_-5^5 (x-sqrt25-x^2)dx=int_-5^5 xdx-int_-5^5 sqrt25-x^2dx$$
Let's calculate the first integral. Obviously it's zero because the function $f(x)=x$ is odd but if you insist you can prove it by summation:
Divide interval from -5 to +5 in $n$ equal segments:
$$Delta x_i=frac10n, quad y_i=x_i=-5+frac 10in,quad i=0, 1, 2,...,n $$
$$A_n=sum_i=0^n-1y_iDelta x_i=sum_i=0^n-1(-5+frac 10in)frac10n$$
$$=-50sum_i=0^n-1frac1n+frac100n^2sum_i=0^n-1i=-50+frac100n^2fracn(n-1)2=-50+50fracn-1n=-frac50n$$
So the first integral is:
$$I_1=lim_nto inftyA_n=lim_nto inftyfrac-50n=0$$
Let's tackle the second one:
$$I_2=int_-5^5 sqrt25-x^2dx$$
The graph of function $f(x)=sqrt25-x^2$ is symmetric with respect to $y$-axis so:
$$I_2=2int_0^5 sqrt25-x^2dx$$
Let's calculate:
$$I_3=int_0^5 sqrt25-x^2dx$$
...by symmation. Again, I will divide the interval of integration into $n$ segments but this time they won't be of equal length. I will introduce variable $varphi$ such that:
$$Deltavarphi=fracpi2n, quad varphi_i=iDeltavarphi=fracipi2n, quad i=0,1,2,...,n$$
$$x_i=5sinvarphi_i$$
$$y_i=sqrt25-x_i^2=5cosvarphi_i$$
$$Delta x_i=x_i+1-x_i=5(sinvarphi_i+1-sinvarphi_i)=5(sin(varphi_i+Deltavarphi)-sinvarphi_i)=5(sinvarphi_i(cosDeltavarphi-1)+cosvarphi_isinDeltavarphi)=5(-2sin^2fracDeltavarphi2sinvarphi_i+cosvarphi_isinDeltavarphi)$$
Having in mind that for small values of $Deltavarphi$:
$$sin Deltavarphi=Deltavarphi+O(Deltavarphi^3)$$
...we can write:
$$Delta x_i=5cosvarphi_iDeltavarphi$$
Now the integral becomes:
$$Delta A_i=y_iDelta x_i=25Deltavarphicos^2varphi_i=frac25Deltavarphi2(1+cos2varphi_i)=frac25Deltavarphi2(1+cosfracipin)$$
$$A_n=sum_i=0^n Delta A_i=frac25Deltavarphi2sum_i=0^n-11+frac25Deltavarphi2sum_i=0^n-1 cosfracipin=frac25nDeltavarphi2+frac25Deltavarphi2sum_i=0^n-1 cosfracipin$$
$$=frac25pi4+frac25pi4nsum_i=0^n-1 cosfracipin$$
By using trigonometric identity of Lagrange you can show that:
$$sum_i=0^n-1 cosfracipin=1+sum_i=1^n-1 cosfracipin=1-frac12+fracsin((n-1+frac12)fracpin)2sinfracpi2n=frac12+fracsin(pi-fracpi2n)2sinfracpi2n=1$$
It means that:
$$A_n=frac25pi4+frac25pi4n$$
$$I_3=lim_ntoinftyA_n=lim_ntoinftyfrac25pi4+frac25pi4n=frac25pi4$$
$$I_2=2I_3=frac25pi2$$
answered Jul 28 at 14:28
Oldboy
2,6001316
You can calcuate these two sums independently:
$$int_-5^5 (x-sqrt25-x^2)dx=int_-5^5 xdx-int_-5^5 sqrt25-x^2dx$$
Let's calculate the first integral. Obviously it's zero because the function $f(x)=x$ is odd but if you insist you can prove it by summation:
Divide interval from -5 to +5 in $n$ equal segments:
$$Delta x_i=frac10n, quad y_i=x_i=-5+frac 10in,quad i=0, 1, 2,...,n $$
$$A_n=sum_i=0^n-1y_iDelta x_i=sum_i=0^n-1(-5+frac 10in)frac10n$$
$$=-50sum_i=0^n-1frac1n+frac100n^2sum_i=0^n-1i=-50+frac100n^2fracn(n-1)2=-50+50fracn-1n=-frac50n$$
So the first integral is:
$$I_1=lim_nto inftyA_n=lim_nto inftyfrac-50n=0$$
Let's tackle the second one:
$$I_2=int_-5^5 sqrt25-x^2dx$$
The graph of function $f(x)=sqrt25-x^2$ is symmetric with respect to $y$-axis so:
$$I_2=2int_0^5 sqrt25-x^2dx$$
Let's calculate:
$$I_3=int_0^5 sqrt25-x^2dx$$
...by symmation. Again, I will divide the interval of integration into $n$ segments but this time they won't be of equal length. I will introduce variable $varphi$ such that:
$$Deltavarphi=fracpi2n, quad varphi_i=iDeltavarphi=fracipi2n, quad i=0,1,2,...,n$$
$$x_i=5sinvarphi_i$$
$$y_i=sqrt25-x_i^2=5cosvarphi_i$$
$$Delta x_i=x_i+1-x_i=5(sinvarphi_i+1-sinvarphi_i)=5(sin(varphi_i+Deltavarphi)-sinvarphi_i)=5(sinvarphi_i(cosDeltavarphi-1)+cosvarphi_isinDeltavarphi)=5(-2sin^2fracDeltavarphi2sinvarphi_i+cosvarphi_isinDeltavarphi)$$
Having in mind that for small values of $Deltavarphi$:
$$sin Deltavarphi=Deltavarphi+O(Deltavarphi^3)$$
...we can write:
$$Delta x_i=5cosvarphi_iDeltavarphi$$
Now the integral becomes:
$$Delta A_i=y_iDelta x_i=25Deltavarphicos^2varphi_i=frac25Deltavarphi2(1+cos2varphi_i)=frac25Deltavarphi2(1+cosfracipin)$$
$$A_n=sum_i=0^n Delta A_i=frac25Deltavarphi2sum_i=0^n-11+frac25Deltavarphi2sum_i=0^n-1 cosfracipin=frac25nDeltavarphi2+frac25Deltavarphi2sum_i=0^n-1 cosfracipin$$
$$=frac25pi4+frac25pi4nsum_i=0^n-1 cosfracipin$$
By using trigonometric identity of Lagrange you can show that:
$$sum_i=0^n-1 cosfracipin=1+sum_i=1^n-1 cosfracipin=1-frac12+fracsin((n-1+frac12)fracpin)2sinfracpi2n=frac12+fracsin(pi-fracpi2n)2sinfracpi2n=1$$
It means that:
$$A_n=frac25pi4+frac25pi4n$$
$$I_3=lim_ntoinftyA_n=lim_ntoinftyfrac25pi4+frac25pi4n=frac25pi4$$
$$I_2=2I_3=frac25pi2$$
answered Jul 28 at 14:28
Oldboy
2,6001316
answered Jul 28 at 14:28
Oldboy
2,6001316
answered Jul 28 at 14:28
Oldboy
2,6001316
answered Jul 28 at 14:28
Oldboy
2,6001316
2,6001316
Thank you that was very thorough. I had to read it several times but I think I got it. Thanks again!
– Fernando Gómez
Jul 28 at 22:31
add a comment |Â
Thank you that was very thorough. I had to read it several times but I think I got it. Thanks again!
– Fernando Gómez
Jul 28 at 22:31
Thank you that was very thorough. I had to read it several times but I think I got it. Thanks again!
– Fernando Gómez
Jul 28 at 22:31
Thank you that was very thorough. I had to read it several times but I think I got it. Thanks again!
– Fernando Gómez
Jul 28 at 22:31
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1
Well, you can use the linearity of the integral and split it into $int x dx$ and $intsqrt25-x^2dx$ and then the second part is even, so you can simplify a bit, doesn't that help?
– Michal Dvořák
Jul 28 at 7:25
For the sum of square roots, see Sum of Square roots formula.
– user539887
Jul 28 at 7:51