Question

$\int\limits^2_1 {3x^2 + 2x +1} \, dx$

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\int\limits^2_1 ({3x^2 + 2x +1} )\, dx = 11}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies\int\limits^2_1 ({3x^2 + 2x +1} )\, dx \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies\int\limits^2_1 ({3x^2 + 2x +1} )\, dx =?$

• According to given question :

$\tt: \implies\int\limits^2_1 ({3x^2 + 2x +1} )\, dx \\ \\ \tt: \implies \int\limits^2_1 \bigg( \frac{3 {x}^{2 + 1} }{2 + 1} + \frac{2 {x}^{1 + 1} }{1 + 1} + \frac{x}{1} \bigg)dx \\ \\ \tt: \implies \bigg[\frac{3 {x}^{3} }{3} + \frac{2x^{2} }{2} + x\bigg]^{2}_{1}\\ \\ \tt: \implies \bigg[ {x}^{3} + {x}^{2} + x\bigg]^{2}_{1} \\ \\ \tt: \implies \bigg[({2}^{3} + {2}^{2} + 2) - ({1}^{3} + {1}^{2} + 1)\bigg] \\ \\ \tt: \implies (8 + 4 + 2) - (1 + 1 + 1) \\ \\ \tt: \implies 14 - 3 \\ \\ \green{\tt: \implies 11} \\ \\ \green{\tt: \implies\int\limits^2_1 ({3x^2 + 2x +1} )\, dx = 11}$

Answer 2

∫(3x² + 2x + 1) dx

By linearity,

= 3∫ x²dx + 2∫ x dx +∫1 dx

$\rule{130}{1.4}$

Solving ∫x²dx :-

Applying power rule :-

∫x²dx = x²⁺¹/2+1

⇒∫x²dx = x³/3

$\rule{130}{1.4}$

Solving ∫x dx :-

Applying power rule :-

∫x dx = x¹⁺¹/1 + 1

⇒∫x dx = x²/2

$\rule{130}{1.4}$

Solving ∫1 dx :-

By constant rule,

∫1 dx = x

$\rule{130}{1.4}$

Now, putting the values in 3∫ x²dx + 2∫ x dx +∫1 dx :-

= 3(x³/3) + 2(x²/2) + x + C

= x³ + x² + x + C            

Now, putting the lower and the upper limits :-

$\sf{[x^3+x^2+x]^2_1}$

$\sf{=(2^3+2^2+2)-(1^3+1^2+1)}\\\\\sf{=(8+4+2)-(1+1+1)}\\\\\sf{=14-3}\\\\\boxed{\sf{=11}}$     ...ANSWER

$\underline{\rule{200}{2}}$

Note :--

• A constant C is always added while doing integration.