Cho hàm số f ( x ) = 2x + 1 . a) ∫ f ( x ) d x = x^2 + x + C .

a) \(\int {f\left( x \right)dx}  = \int {\left( {2x + 1} \right)dx}  = {x^2} + x + C\).

b) \(\int {\left( {x - 1} \right)f\left( x \right)dx} \)\( = \int {\left( {x - 1} \right)\left( {2x + 1} \right)dx} \)\( = \int {\left( {2{x^2} - x - 1} \right)dx} \)\( = \frac{2}{3}{x^3} - \frac{{{x^2}}}{2} - x + C\).

c) \(G\left( x \right) = {x^2} + x + C\). Có \(G\left( 2 \right) = 6 + C = 5 \Rightarrow C =  - 1\).

Do đó \(G\left( x \right) = {x^2} + x - 1\).

d) \(F\left( x \right) = {x^2} + x + C\). Có \(F\left( 1 \right) = 2 + C = 2 \Rightarrow C = 0\).

Do đó \(F\left( x \right) = {x^2} + x = x\left( {x + 1} \right)\).

Khi đó \(F\left( 1 \right) = 1.2;F\left( 2 \right) = 2.3;...;F\left( {99} \right) = 99.100;F\left( {100} \right) = 100.101\).

Khi đó \(\frac{1}{{F\left( 1 \right)}} + \frac{1}{{F\left( 2 \right)}} + ... + \frac{1}{{F\left( {99} \right)}} + \frac{1}{{F\left( {100} \right)}} = \frac{1}{{1.2}} + \frac{1}{{2.3}} + ... + \frac{1}{{99.100}} + \frac{1}{{100.101}}\)

\( = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{{99}} - \frac{1}{{100}} + \frac{1}{{100}} - \frac{1}{{101}} = 1 - \frac{1}{{101}} = \frac{{100}}{{101}}\).

Suy ra \(a = 100;b = 101\). Do đó \(a + b = 201\).

Đáp án: a) Đúng;b) Đúng;   c) Đúng;d) Đúng.