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මේ A/L ගාණ හදල දියල්ලකො
- Thread starter elaelkiri
- Start date
Full solution... Thanks to Wolfram Alpha
Compute the definite integral:
integral_1^8 1/(x^(3/2) + x^(4/3)) dx
For the integrand 1/(x^(3/2) + x^(4/3)), substitute u = x^(1/6) and du = 1/(6 x^(5/6)) dx.
This gives a new lower bound u = 1^(1/6) = 1 and upper bound u = 8^(1/6) = sqrt(2):
= 6 integral_1^sqrt(2) u^5/(u^9 + u^8) du
For the integrand u^5/(u^9 + u^8), cancel common terms in the numerator and denominator:
= 6 integral_1^sqrt(2) 1/(u^3 (u + 1)) du
For the integrand 1/(u^3 (u + 1)), use partial fractions:
= 6 integral_1^sqrt(2)-1/(u + 1) + 1/u - 1/u^2 + 1/u^3 du
Integrate the sum term by term and factor out constants:
= -6 integral_1^sqrt(2) 1/(u + 1) du + 6 integral_1^sqrt(2) 1/u du - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
For the integrand 1/(u + 1), substitute s = u + 1 and ds = du.
This gives a new lower bound s = 1 + 1 = 2 and upper bound s = 1 + sqrt(2):
= -6 integral_2^(1 + sqrt(2)) 1/s ds + 6 integral_1^sqrt(2) 1/u du - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
Apply the fundamental theorem of calculus.
The antiderivative of 1/s is log(s):
= (-6 log(s)) right bracketing bar _2^(1 + sqrt(2)) + 6 integral_1^sqrt(2) 1/u du - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
Evaluate the antiderivative at the limits and subtract.
(-6 log(s)) right bracketing bar _2^(1 + sqrt(2)) = (-6 log(1 + sqrt(2))) - (-6 log(2)) = 6 log(2 (sqrt(2) - 1)):
= 6 log(2 (sqrt(2) - 1)) + 6 integral_1^sqrt(2) 1/u du - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u):
= 6 log(2 (sqrt(2) - 1)) + 6 log(u) right bracketing bar _1^sqrt(2) - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
Evaluate the antiderivative at the limits and subtract.
6 log(u) right bracketing bar _1^sqrt(2) = 6 log(sqrt(2)) - 6 log(1) = log(8):
= log(8) + 6 log(2 (sqrt(2) - 1)) - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
Apply the fundamental theorem of calculus.
The antiderivative of 1/u^2 is -1/u:
= log(8) + 6 log(2 (sqrt(2) - 1)) + 6/u right bracketing bar _1^sqrt(2) + 6 integral_1^sqrt(2) 1/u^3 du
Evaluate the antiderivative at the limits and subtract.
6/u right bracketing bar _1^sqrt(2) = 6/sqrt(2) - 6 1/1 = 3 (sqrt(2) - 2):
= 3 (sqrt(2) - 2) + log(8) + 6 log(2 (sqrt(2) - 1)) + 6 integral_1^sqrt(2) 1/u^3 du
Apply the fundamental theorem of calculus.
The antiderivative of 1/u^3 is -1/(2 u^2):
= 3 (sqrt(2) - 2) + log(8) + 6 log(2 (sqrt(2) - 1)) + (-3/u^2) right bracketing bar _1^sqrt(2)
Evaluate the antiderivative at the limits and subtract.
(-3/u^2) right bracketing bar _1^sqrt(2) = (-3/sqrt(2)^2) - (-3/1^2) = 3/2:
= 3/2 + 3 (sqrt(2) - 2) + log(8) + 6 log(2 (sqrt(2) - 1))
Which is equal to:
Answer: |
| = -9/2 + 3 sqrt(2) + log(512) - 6 log(1 + sqrt(2)) = 0.6927
Compute the definite integral:
integral_1^8 1/(x^(3/2) + x^(4/3)) dx
For the integrand 1/(x^(3/2) + x^(4/3)), substitute u = x^(1/6) and du = 1/(6 x^(5/6)) dx.
This gives a new lower bound u = 1^(1/6) = 1 and upper bound u = 8^(1/6) = sqrt(2):
= 6 integral_1^sqrt(2) u^5/(u^9 + u^8) du
For the integrand u^5/(u^9 + u^8), cancel common terms in the numerator and denominator:
= 6 integral_1^sqrt(2) 1/(u^3 (u + 1)) du
For the integrand 1/(u^3 (u + 1)), use partial fractions:
= 6 integral_1^sqrt(2)-1/(u + 1) + 1/u - 1/u^2 + 1/u^3 du
Integrate the sum term by term and factor out constants:
= -6 integral_1^sqrt(2) 1/(u + 1) du + 6 integral_1^sqrt(2) 1/u du - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
For the integrand 1/(u + 1), substitute s = u + 1 and ds = du.
This gives a new lower bound s = 1 + 1 = 2 and upper bound s = 1 + sqrt(2):
= -6 integral_2^(1 + sqrt(2)) 1/s ds + 6 integral_1^sqrt(2) 1/u du - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
Apply the fundamental theorem of calculus.
The antiderivative of 1/s is log(s):
= (-6 log(s)) right bracketing bar _2^(1 + sqrt(2)) + 6 integral_1^sqrt(2) 1/u du - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
Evaluate the antiderivative at the limits and subtract.
(-6 log(s)) right bracketing bar _2^(1 + sqrt(2)) = (-6 log(1 + sqrt(2))) - (-6 log(2)) = 6 log(2 (sqrt(2) - 1)):
= 6 log(2 (sqrt(2) - 1)) + 6 integral_1^sqrt(2) 1/u du - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u):
= 6 log(2 (sqrt(2) - 1)) + 6 log(u) right bracketing bar _1^sqrt(2) - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
Evaluate the antiderivative at the limits and subtract.
6 log(u) right bracketing bar _1^sqrt(2) = 6 log(sqrt(2)) - 6 log(1) = log(8):
= log(8) + 6 log(2 (sqrt(2) - 1)) - 6 integral_1^sqrt(2) 1/u^2 du + 6 integral_1^sqrt(2) 1/u^3 du
Apply the fundamental theorem of calculus.
The antiderivative of 1/u^2 is -1/u:
= log(8) + 6 log(2 (sqrt(2) - 1)) + 6/u right bracketing bar _1^sqrt(2) + 6 integral_1^sqrt(2) 1/u^3 du
Evaluate the antiderivative at the limits and subtract.
6/u right bracketing bar _1^sqrt(2) = 6/sqrt(2) - 6 1/1 = 3 (sqrt(2) - 2):
= 3 (sqrt(2) - 2) + log(8) + 6 log(2 (sqrt(2) - 1)) + 6 integral_1^sqrt(2) 1/u^3 du
Apply the fundamental theorem of calculus.
The antiderivative of 1/u^3 is -1/(2 u^2):
= 3 (sqrt(2) - 2) + log(8) + 6 log(2 (sqrt(2) - 1)) + (-3/u^2) right bracketing bar _1^sqrt(2)
Evaluate the antiderivative at the limits and subtract.
(-3/u^2) right bracketing bar _1^sqrt(2) = (-3/sqrt(2)^2) - (-3/1^2) = 3/2:
= 3/2 + 3 (sqrt(2) - 2) + log(8) + 6 log(2 (sqrt(2) - 1))
Which is equal to:
Answer: |
| = -9/2 + 3 sqrt(2) + log(512) - 6 log(1 + sqrt(2)) = 0.6927
You will not be able to see the links.. Only available in the paid versions of Wolfram Alpha.
But you can download the pdf from here, Hope it helps
PDF වුණාට කමක් නෑ. ඔය තියෙන්නෙ ගින්දර වගේ
thanks
apoo thopila oya thama AL s kerena unda?? deyyak kiyannada math maths kale..mata dan 30 ta kittui....mata C maths welata AL...
oya ganan hadanakota oka haduwama hithei shaa mama meka haduwa mata kisi deyak dan prashna nemai..anith ewa simple kiyala wage deyak oluwata ei...eth jiwithe prashna owa nemai machan...to be continued
oya ganan hadanakota oka haduwama hithei shaa mama meka haduwa mata kisi deyak dan prashna nemai..anith ewa simple kiyala wage deyak oluwata ei...eth jiwithe prashna owa nemai machan...to be continued
Photomath android app eka try karala balahn.
Uththare denwa uba gana cam ekata allapu gaman.
Oyawage ewata harida danne. Mata owa therenne ne bn.
Try karala balahn. Hadapu steps ekkama enw.
Uththare denwa uba gana cam ekata allapu gaman.
Oyawage ewata harida danne. Mata owa therenne ne bn.
Try karala balahn. Hadapu steps ekkama enw.
You are correct. Photomath is able to solve some integrals. But it's no comparison to Wolfram. But these are quite simple integrals and easily solvable without anything. The problem is how to post it in this text based forum the steps involved. The basic premise is to know how the integration is done.. not to use a software tool.
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