Question

❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​

Answer 1

${\boxed{\boxed{\begin{array}{cc}\bf \: \to \:given : \\ \\ \odot \: \displaystyle \int_ {0}^{3} \rm \:f(x) \: dx = 6 \: \: \: - - - (1) \\ \\ \odot \: \displaystyle \int_ {3}^{5} \rm \:f(x) \: dx = 4 \: \: \: - - - (2) \\ \\ \blue{ \underline{ \sf \: we \: have \: to \: find : \: }} \\ \\ \displaystyle \int_ {0}^{5} \rm \:(3 + 2 \: f(x)) \: dx = \: ?\end{array}}}}$

${\boxed{\boxed{\begin{array}{cc} \red{ \underline{\bf \: solution}} \\ \\ \displaystyle \int_ {0}^{5} \rm \:(3 + 2 \: f(x)) \: dx \\ \\ = \displaystyle \int_ {0}^{5} \rm \:3 \: dx + \displaystyle \int_ {0}^{5} \rm \:2 \: f(x) \: dx \\ \\ = { \huge{[}}3x{ \huge{]}}_0^5 + 2\displaystyle \int_ {0}^{5} \rm \:f(x) \: dx \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: \to \:we \: know : \\ \\ \displaystyle \int_ {a}^{c} \rm \:f(x) \: dx = \displaystyle \int_ {a}^{b} \rm \:f(x) \: dx + \displaystyle \int_ {b}^{c} \rm \:f(x) \: dx\end{array}}}} \\ \\ = (3 \times 5 - 3 \times 0) + 2 \left(\displaystyle \int_ {0}^{3} \rm \:f(x) \: dx + \displaystyle \int_ {3}^{5} \rm \:f(x) \: dx \right) \\ \\ = 15 + 2(6 + 4) \\ \\ \orange{ \boxed{ \bf \: by \: eqn.(1) \: \: and \: (2) }} \\ \\ = 15 + 2 \times 10 \\ \\ = 15 + 20 \\ \\ = 35\end{array}}}}$

$\orange{ \boxed{ \therefore \displaystyle \int_ {0}^{5} \rm \:(3 + 2 \: f(x)) \: dx = 35}}$

Answer 2

Given Question :-

$If \: \displaystyle\int_0^3\sf f(x)dx = 6 \: and \: \displaystyle\int_3^5\sf f(x)dx = 4, \: then \: \displaystyle\int_0^5\sf (3 + 2f(x))dx = - - -$

$\green{\large\underline{\sf{Solution-}}}$

Given that,

$\rm :\longmapsto\: \displaystyle\int_0^3\sf f(x)dx = 6 \:$

and

$\rm :\longmapsto\:\displaystyle\int_3^5\sf f(x)dx = 4$

Consider,

$\rm :\longmapsto\:\displaystyle\int_0^5\sf (3 + 2f(x)) \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int_0^5\sf 3 \: dx + 2\displaystyle\int_0^5\sf f(x) \: dx$

$\rm \: = \:3 \displaystyle\int_0^5\sf \: dx + 2\bigg(\displaystyle\int_0^3\sf f(x) \: dx + \displaystyle\int_3^5\sf f(x)dx\bigg)$

We know,

$\boxed{ \tt{ \: \displaystyle\int\sf \: k \: dx \: = \: kx \: + \: c \: }}$

So, using this, we get

$\rm \: = \: 3\bigg(x \bigg) _0^5\sf + 2(6 + 4)$

$\rm \: = \: 3(5 - 0) + 2(10)$

$\rm \: = \: 15 + 20$

$\rm \: = \: 35$

Hence,

$\purple{\rm \implies\:\boxed{ \tt{ \: \displaystyle\int_0^5\sf (3 + 2f(x)) \: dx \: = \: 35 \: }}}$

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Additional Information :-

$\pink{\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}}$

Properties of Definite integrals

$\boxed{ \tt{ \: \displaystyle\int_a^b\sf f(x) dx \: = \: \displaystyle\int_a^b\sf f(y) dy \: }}$

$\boxed{ \tt{ \: \displaystyle\int_a^b\sf f(x) dx \: = \: - \: \displaystyle\int_b^a\sf f(x) dx \: }}$

$\boxed{ \tt{ \: \displaystyle\int_0^a\sf f(x) dx \: = \: \displaystyle\int_0^a\sf f(a - x) dx \: }}$

$\boxed{ \tt{ \: \displaystyle\int_a^b\sf f(x) dx \: = \: \displaystyle\int_a^b\sf f(a + b- x) dx \: }}$

$\boxed{ \tt{ \: \displaystyle\int_0^{2a}\sf f(x) dx \: = 2 \: \displaystyle\int_0^a\sf f(x) dx \:if \: f(2a - x) = f(x) }}$

$\boxed{ \tt{ \: \displaystyle\int_0^{2a}\sf f(x) dx \: = 0 \:if \: f(2a - x) = - f(x) }}$