Question

solve the question of integration of class 11th

Answer 1

Heyy !
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$\int\limits^2_1 {(4x^3-5x^2+6x+9)} \, dx$

as we know , 

$\int\limits {(u+v)} \, dx =\int (u )dx + \int (v) dx$
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$\int (4x^3) dx - \int (5x^2) dx + \int(6x) dx +\int 9dx$

$\frac{4x^4}{4} - \frac{5x^3}{3} + \frac{6x^2}{2} + 9x$

$\frac{6x^4 - 10 x^3 + 18x^2 + 54 x}{6}$

$\frac{3x^4 -5x^3 +9x^2 + 27 x}{3}$ + C
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Definite integration :
put the values of x = 2 and x = 1 
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x = 2 

$\frac{48 - 40 + 36 + 54}{3}$

$\frac{98}{3}$
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x = 1 
= 34/3 
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Definite integration = 98/3 - 34/3 
                                    = 64/3
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Answer 2

Hey there !!!!

$\int\limits^2_1 {4x^3-5x^2+6x+9 } \, dx$ --- Equation 1

Integrating by splitting equation 1

$\int\limits^2_1 {4x^3} \, dx - \int\limits^2_1 {5x^2} \, dx + \int\limits^2_1 {6x} \, dx + \int\limits^2_1 {9} \, dx$

=($(4x^4/4 -5x^3/3 +6x^2/2+9x) ^{2} _{1}$ ---- Equation 2
Substituting lower and upper limits in equation 2

=( (2)⁴-5*(2)³/3+3(2)²+18)-(1-5/3+3+9)

=16-40/3+12+18-1+5/3-12

=33-40/3+5/3

=33-35/3

=(33*3-35)/3=(99-35)/3=64/3

Hope this helped you !!!!