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Christoph J. Scherr 2024-09-09 13:22:23 +02:00
parent 02145437cd
commit 4f8d02528b
Signed by: cscherrNT
GPG Key ID: 8E2B45BC51A27EA7
2 changed files with 25 additions and 18 deletions

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@ -148,23 +148,23 @@ $
Calculate the following integrals using substitution:
#set enum(numbering: "(a)")
+ $ integral (2x+7)/(x^2+7x+3) d x $
+ $ integral (cos(ln(x)))/(x) d x $
+ $ integral (2x+7)/(x^2+7x+3) thin d x $
+ $ integral (cos(ln(x)))/(x) thin d x $
])
#set enum(numbering: "(a)")
+ $
A &= integral (2x+7)/(x^2+7x+3) d x \
&= integral (a'(x))/(a(x)) d x \
&= integral (a'(x))/(a(x)) d x \
&= integral 1/(a(x)) dot a'(x) d x; wide d u = a'(x) d x \
A &= integral (2x+7)/(x^2+7x+3) thin d x \
&= integral (a'(x))/(a(x)) thin d x \
&= integral (a'(x))/(a(x)) thin d x \
&= integral 1/(a(x)) dot a'(x) thin d x; wide d u = a'(x) thin d x \
&= integral 1/u d u \
&= ln(u) = ln(x^2+7x+3) checkmark
$
+ $
A &= integral (cos(ln(x)))/(x) d x \
&= integral cos(ln(x)) dot 1/x d x \
&= integral cos(u) dot u' d x \
A &= integral (cos(ln(x)))/(x) thin d x \
&= integral cos(ln(x)) dot 1/x thin d x \
&= integral cos(u) dot u' thin d x \
&= integral cos(u) dot d u \
&= sin(u) \
&= sin(ln(x)) \
@ -180,16 +180,23 @@ $
Calculate the following integrals using partial integration:
#set enum(numbering: "(a)")
+ $ integral (x+2) dot e^(2x) d x $
+ $ integral x^2 dot ln(x) d x $
+ $ integral (x+2) dot e^(2x) thin d x $
+ $ integral x^2 dot ln(x) thin d x $
])
#set enum(numbering: "(a)")
+ $
A &= integral (x+2) dot e^(2x) d x \
&= integral u(x) dot v'(x) d x \
&= u(x) dot v(x) - integral u'(x) dot v(x) d x \
&= (x+2) dot (e^(2x) dot 1/2x) - integral e^(2x) d x \
&= (x+2) dot (e^(2x) dot 1/2x) - e^(2x) dot 1/2x \
&= x dot (e^(2x) dot 1/2x) + e^(2x) dot 1/2x \
$
u(x) &= x => u'(x) = 1; space v(x) = e^(2x) dot 1/2 x => v'(x) = e^(2x) \
A &= integral (x+2) dot e^(2x) thin d x \
&= integral x dot e^(2x) thin d x + integral 2 dot e^(2x) thin d x wide \
&= integral x dot e^(2x) thin d x + 2 dot integral e^(2x) thin d x \
&= integral x dot e^(2x) thin d x + 2 dot 1/2 e^(2x) \
&= integral x dot e^(2x) thin d x + e^(2x) \
&= integral u(x) dot v'(x) thin d x + e^(2x) \
&= [u(x) dot v(x)] - integral u'(x) dot v(x) thin d x + e^(2x) \
&= x dot e^(2x) dot 1/2 - integral 1 dot e^(2x) dot 1/2 thin d x + e^(2x) \
&= 1/2 x dot e^(2x) - e^(2x) 1/4 + e^(2x) \
&= e^(2x) dot (1/2 x - 1/4 + 1) \
&= 1/2 e^(2x) dot (x - 1/2 + 2) \
&= 1/2 e^(2x) dot (x + 3/2) checkmark
| $