Question

Evaluate the following:
$\displaystyle \rm 1) \int(2x + 3 {)}^{2} dx$
$\displaystyle \rm 2)\int_{0}^{2} (2x + 3 {)}^{2} dx$
Please solve this :)
Thanks!​

Answer 1

Solution 1:

The given integral is:

$=\displaystyle \rm\int {(2x + 3)}^{2} \: dx$

Can be written as:

$=\displaystyle \rm\int(4 {x}^{2} + 12x + 9)\: dx$

$=\displaystyle \rm\int4 {x}^{2} \: dx + \int 12x \: dx+ \int9\: dx$

$=\displaystyle \rm4\int{x}^{2} \: dx + 12\int x \: dx+ 9\int dx$

$=\displaystyle \rm4 \dfrac{ {x}^{3} }{3} + 12 \dfrac{ {x}^{2} }{2} + 9x$

$=\displaystyle \rm\dfrac{ 4{x}^{3} }{3} +\dfrac{12 {x}^{2} }{2} + 9x$

$=\displaystyle \rm\dfrac{ 4{x}^{3} }{3} +6{x}^{2} + 9x + C$

Therefore:

$\displaystyle \rm \longrightarrow\int {(2x + 3)}^{2} \: dx = \dfrac{4 {x}^{3} }{3} + 6 {x}^{2} + 9x + C$

Solution 2:

The given definite integral is:

$\displaystyle \rm = \int_{0}^{2} (2x + 3 {)}^{2} dx$

$=\displaystyle \rm \bigg(\dfrac{ 4{x}^{3} }{3} +6{x}^{2} + 9x \bigg) \bigg| ^{2}_{0}$

$=\displaystyle \rm \dfrac{ 4 \times {2}^{3} }{3} +6 \times {2}^{2} + 9 \times 2 -0$

$=\displaystyle \rm \dfrac{32}{3} + 24 + 18$

$=\displaystyle \rm10 \dfrac{2}{3} + 42$

$=\displaystyle \rm52 \dfrac{2}{3}$

★ Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$