Definite integral finding unknown
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Given ∫1^2 f(u) du =-5, ∫1^2 h(u) du =4, ∫2^5 f(u) du =8.
Find the value of p if ∫1^5 [f(u)-3pu] du=39
I just know few on how to solve this
39 = ∫1^5 f(u) du - ∫1^5 3pu du
39 = ∫1^2 f(u) du + ∫2^5 f(u) du - ∫1^5 pu du
39 = -5+8 -3∫1^5 pu du
36 = -3∫1^5 pu du
What to do next?
definite-integrals
asked Dec 3 '18 at 18:13
A IzaliaA Izalia
11
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add a comment |
$begingroup$
Given ∫1^2 f(u) du =-5, ∫1^2 h(u) du =4, ∫2^5 f(u) du =8.
Find the value of p if ∫1^5 [f(u)-3pu] du=39
I just know few on how to solve this
39 = ∫1^5 f(u) du - ∫1^5 3pu du
39 = ∫1^2 f(u) du + ∫2^5 f(u) du - ∫1^5 pu du
39 = -5+8 -3∫1^5 pu du
36 = -3∫1^5 pu du
What to do next?
definite-integrals
asked Dec 3 '18 at 18:13
A IzaliaA Izalia
11
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Oh and the ∫5/1 [f(u)-3pu] du=39
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– A Izalia
Dec 3 '18 at 18:15
1
$begingroup$
Please use MathJax for clarity. For example, by ∫2/1 f(u) du, do you mean $int_1^2 f(u) du$? If so, writeint_1^2 f(u) du.
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– J.G.
Dec 3 '18 at 18:17
add a comment |
$begingroup$
Given ∫1^2 f(u) du =-5, ∫1^2 h(u) du =4, ∫2^5 f(u) du =8.
Find the value of p if ∫1^5 [f(u)-3pu] du=39
I just know few on how to solve this
39 = ∫1^5 f(u) du - ∫1^5 3pu du
39 = ∫1^2 f(u) du + ∫2^5 f(u) du - ∫1^5 pu du
39 = -5+8 -3∫1^5 pu du
36 = -3∫1^5 pu du
What to do next?
definite-integrals
asked Dec 3 '18 at 18:13
A IzaliaA Izalia
11
$endgroup$
Given ∫1^2 f(u) du =-5, ∫1^2 h(u) du =4, ∫2^5 f(u) du =8.
Find the value of p if ∫1^5 [f(u)-3pu] du=39
I just know few on how to solve this
39 = ∫1^5 f(u) du - ∫1^5 3pu du
39 = ∫1^2 f(u) du + ∫2^5 f(u) du - ∫1^5 pu du
39 = -5+8 -3∫1^5 pu du
36 = -3∫1^5 pu du
What to do next?
definite-integrals
definite-integrals
asked Dec 3 '18 at 18:13
A IzaliaA Izalia
11
asked Dec 3 '18 at 18:13
A IzaliaA Izalia
11
edited Dec 3 '18 at 18:29
A Izalia
asked Dec 3 '18 at 18:13
A IzaliaA Izalia
11
asked Dec 3 '18 at 18:13
A IzaliaA Izalia
11
asked Dec 3 '18 at 18:13
A IzaliaA Izalia
11
11
$begingroup$
Oh and the ∫5/1 [f(u)-3pu] du=39
$endgroup$
– A Izalia
Dec 3 '18 at 18:15
1
$begingroup$
Please use MathJax for clarity. For example, by ∫2/1 f(u) du, do you mean $int_1^2 f(u) du$? If so, writeint_1^2 f(u) du.
$endgroup$
– J.G.
Dec 3 '18 at 18:17
add a comment |
$begingroup$
Oh and the ∫5/1 [f(u)-3pu] du=39
$endgroup$
– A Izalia
Dec 3 '18 at 18:15
1
$begingroup$
Please use MathJax for clarity. For example, by ∫2/1 f(u) du, do you mean $int_1^2 f(u) du$? If so, writeint_1^2 f(u) du.
$endgroup$
– J.G.
Dec 3 '18 at 18:17
$begingroup$
Oh and the ∫5/1 [f(u)-3pu] du=39
$endgroup$
– A Izalia
Dec 3 '18 at 18:15
$begingroup$
Oh and the ∫5/1 [f(u)-3pu] du=39
$endgroup$
– A Izalia
Dec 3 '18 at 18:15
1
1
$begingroup$
Please use MathJax for clarity. For example, by ∫2/1 f(u) du, do you mean $int_1^2 f(u) du$? If so, write
int_1^2 f(u) du.$endgroup$
– J.G.
Dec 3 '18 at 18:17
$begingroup$
Please use MathJax for clarity. For example, by ∫2/1 f(u) du, do you mean $int_1^2 f(u) du$? If so, write
int_1^2 f(u) du.$endgroup$
– J.G.
Dec 3 '18 at 18:17
add a comment |
1 Answer
1
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Assuming I've understood your question:
Since $int_1^5 f(u) du=-5+8=3$, $3p=frac{3-39}{int_1^5 u du}=-frac{36}{12}=-3$, i.e. $p=-1$.
answered Dec 3 '18 at 18:19
J.G.J.G.
27.7k22843
$endgroup$
$begingroup$
Thankyouuu so much
$endgroup$
– A Izalia
Dec 3 '18 at 18:37
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Assuming I've understood your question:
Since $int_1^5 f(u) du=-5+8=3$, $3p=frac{3-39}{int_1^5 u du}=-frac{36}{12}=-3$, i.e. $p=-1$.
answered Dec 3 '18 at 18:19
J.G.J.G.
27.7k22843
$endgroup$
$begingroup$
Thankyouuu so much
$endgroup$
– A Izalia
Dec 3 '18 at 18:37
add a comment |
$begingroup$
Assuming I've understood your question:
Since $int_1^5 f(u) du=-5+8=3$, $3p=frac{3-39}{int_1^5 u du}=-frac{36}{12}=-3$, i.e. $p=-1$.
answered Dec 3 '18 at 18:19
J.G.J.G.
27.7k22843
$endgroup$
$begingroup$
Thankyouuu so much
$endgroup$
– A Izalia
Dec 3 '18 at 18:37
add a comment |
$begingroup$
Assuming I've understood your question:
Since $int_1^5 f(u) du=-5+8=3$, $3p=frac{3-39}{int_1^5 u du}=-frac{36}{12}=-3$, i.e. $p=-1$.
answered Dec 3 '18 at 18:19
J.G.J.G.
27.7k22843
$endgroup$
Assuming I've understood your question:
Since $int_1^5 f(u) du=-5+8=3$, $3p=frac{3-39}{int_1^5 u du}=-frac{36}{12}=-3$, i.e. $p=-1$.
answered Dec 3 '18 at 18:19
J.G.J.G.
27.7k22843
answered Dec 3 '18 at 18:19
J.G.J.G.
27.7k22843
answered Dec 3 '18 at 18:19
J.G.J.G.
27.7k22843
answered Dec 3 '18 at 18:19
J.G.J.G.
27.7k22843
27.7k22843
$begingroup$
Thankyouuu so much
$endgroup$
– A Izalia
Dec 3 '18 at 18:37
add a comment |
$begingroup$
Thankyouuu so much
$endgroup$
– A Izalia
Dec 3 '18 at 18:37
$begingroup$
Thankyouuu so much
$endgroup$
– A Izalia
Dec 3 '18 at 18:37
$begingroup$
Thankyouuu so much
$endgroup$
– A Izalia
Dec 3 '18 at 18:37
add a comment |
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$begingroup$
Oh and the ∫5/1 [f(u)-3pu] du=39
$endgroup$
– A Izalia
Dec 3 '18 at 18:15
1
$begingroup$
Please use MathJax for clarity. For example, by ∫2/1 f(u) du, do you mean $int_1^2 f(u) du$? If so, write
int_1^2 f(u) du.$endgroup$
– J.G.
Dec 3 '18 at 18:17