Question

If y = 3x² then find area under y - x curve inbetween x = 1 to x = 2
a) 6 unit b) 7 unit c) 8 unit d) 9 unit​

Answer 1

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

Givemn curve is

$\rm :\longmapsto\:y = {3x}^{2}$

We know,

Its represents a upper parabola having vertex (0, 0).

So, Required area between the curve with respect to x - axis is

$\rm \:  =  \:  \: \displaystyle\int_{1}^2 \rm y \: dx$

$\rm \:  =  \:  \: \displaystyle\int_{1}^2 \rm {3x}^{2} \: dx$

$\rm \:  =  \: 3 \: \displaystyle\int_{1}^2 \rm {x}^{2} \: dx$

We know that,

$\underbrace{\boxed{ \tt{\displaystyle\int \rm \: {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} + c }}}$

So, using this,

$\rm \:  =  \: 3 \: \bigg(\dfrac{ {x}^{2 + 1} }{2 + 1} \bigg) _{1}^2$

$\rm \:  =  \: 3 \: \bigg(\dfrac{ {x}^{3} }{3} \bigg) _{1}^2$

$\rm \:  =  \:  \: \bigg( {x}^{3} \bigg) _{1}^2$

$\rm \:  =  \:  \: {2}^{3} - {1}^{3}$

$\rm \:  =  \:  \: 8 - 1$

$\rm \:  =  \:  \: 7 \: sq. \: units$

Additional Information:-

$\underbrace{\boxed{ \tt{\displaystyle\int_{a}^b \rm f(x)dx \: = \: \displaystyle\int_{a}^b \rm f(y)dy }}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{a}^b \rm f(x)dx \: = - \: \displaystyle\int_{b}^a \rm f(x)dx }}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{a}^b \rm f(x)dx \: = \: \displaystyle\int_{a}^b \rm f(a + b - x)dx }}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{0}^b \rm f(x)dx \: = \: \displaystyle\int_{0}^b \rm f(b - x)dx }}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{ - b}^b \rm f(x)dx \: = 2 \: \displaystyle\int_{0}^b \rm f(x)dx \: if \: f( - x) = f(x)}}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{ - b}^b \rm f(x)dx \: = 0\: if \: f( - x) = - f(x)}}}$