fill in the blanks. A nomedic tribe live in bedouins is called __________
Answers
\large\underline{\sf{Solution-}}
Solution−
Given integral is
\rm \: \displaystyle \int \: \rm\frac{ {2}^{x + 1} \: - \: {5}^{x - 1} }{ {10}^{x} } dx∫
10
x
2
x+1
−5
x−1
dx
can be rewritten as
\rm \: = \: \displaystyle \int \: \rm\frac{2. {2}^{x} \: - \: {5}^{x}. {5}^{ - 1} }{ {10}^{x} } dx= ∫
10
x
2.2
x
−5
x
.5
−1
dx
\begin{gathered}\rm \: = \: 2\displaystyle \int\rm \: \frac{ {2}^{x} }{ {10}^{x} } \: dx \: - \: \frac{1}{5}\displaystyle \int\rm \: \frac{ {5}^{x} }{ {10}^{x} } \: dx \\ \end{gathered}
= 2∫
10
x
2
x
dx−
5
1
∫
10
x
5
x
dx
\begin{gathered}\rm \: = \: 2\displaystyle \int\rm \: {\bigg(\dfrac{2}{10} \bigg) }^{x} \: dx \: - \: \frac{1}{5}\displaystyle \int\rm \: {\bigg(\dfrac{5}{10} \bigg) }^{x} \: dx \\ \end{gathered}
= 2∫(
10
2
)
x
dx−
5
1
∫(
10
5
)
x
dx
\begin{gathered}\rm \: = \: 2\displaystyle \int\rm \: {\bigg(\dfrac{1}{5} \bigg) }^{x} \: dx \: - \: \frac{1}{5}\displaystyle \int\rm \: {\bigg(\dfrac{1}{2} \bigg) }^{x} \: dx \\ \end{gathered}
= 2∫(
5
1
)
x
dx−
5
1
∫(
2
1
)
x
dx
We know,
\begin{gathered}\boxed{\sf{ \:\displaystyle \int\rm \: {a}^{x} \: dx \: = \: \frac{ {a}^{x} }{loga} + c \: \: }} \\ \end{gathered}
∫a
x
dx=
loga
a
x
+c
So, using this result, we get
\begin{gathered}\rm \: = \: 2 \times \dfrac{{\bigg(\dfrac{1}{5} \bigg) }^{x}}{log\dfrac{1}{5} } \: - \: \dfrac{1}{5} \times \dfrac{{\bigg(\dfrac{1}{2} \bigg) }^{x}}{log\dfrac{1}{2} } + c \\ \end{gathered}
= 2×
log
5
1
(
5
1
)
x
−
5
1
×
log
2
1
(
2
1
)
x
+c
can be further simplified to
\begin{gathered}\rm \: = \: 2 \times \dfrac{ {5}^{ - x} }{log {5}^{ - 1} } \: - \: \dfrac{1}{5} \times \dfrac{ {2}^{ - x} }{log {2}^{ - 1} } + c \\ \end{gathered}
= 2×
log5
−1
5
−x
−
5
1
×
log2
−1
2
−x
+c
\begin{gathered}\rm \: = \: - 2 \: \dfrac{ {5}^{ - x} }{log5} \: + \: \dfrac{1}{5} \: \dfrac{ {2}^{ - x} }{log 2} + c \\ \end{gathered}
= −2
log5
5
−x
+
5
1
log2
2
−x
+c
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
Additional Information
\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}
f(x)
k
sinx
cosx
sec
2
x
cosec
2
x
secxtanx
cosecxcotx
tanx
x
1
e
x
∫f(x)dx
kx+c
−cosx+c
sinx+c
tanx+c
−cotx+c
secx+c
−cosecx+c
logsecx+c
logx+c
e
x
+c
Answer:
dikus bro
Explanation:
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