Question

$\tt{Question}$​

Answer 1

Answer:

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

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

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

${\blue{\rule{25pt}{1pt}{\pink{\rule{30pt}{3pt}{\red{\rule{100pt}{5pt}{\blue{\rule{25pt}{1pt}{\pink{\rule{30pt}{3pt} }}}}}}}}}}$

Answer 2

$\huge\sf\fbox\purple{ ★ Solution ★}$

°°° Explanation °°°

To find zero of the polynomial, p(x)=0

(i) If p(x)=x+5=0 then x=−5, i.e. −5 is the zero.

(ii) If p(x)=x−5=0 then x=5, i.e. 5 is the zero.

(iii) If p(x)=2x+5=0 then x= 2

−5 , i.e. 2−5 is the zero.

(iv) If p(x)=3x−2=0 then x= 32

, i.e. 32 is the zero.

(v) If p(x)=3x=0 then x=0, i.e. 0 is the zero.

(vi) If p(x)=ax=0 then x=0, i.e. 0 is the zero.

(vii) If p(x)=cx+d=0 then x= c−d

, i.e. c −d

is the zero.