22M.2.SL.TZ2.2

METHOD 1

recognises that $g\left(x\right)=\int \left(3x^2+5{\text{e}}^x\right)dx$           (M1)

$g\left(x\right)=x^3+5{\text{e}}^x\left(+C\right)$           (A1)(A1)

Note: Award A1 for each integrated term.

 

substitutes $x=0$ and $y=4$ into their integrated function (must involve $+C$)           (M1)

$4=0+5+C\Rightarrow C=-1$

$g\left(x\right)=x^3+5{\text{e}}^x-1$           A1

 

METHOD 2

attempts to write both sides in the form of a definite integral           (M1)

$\underset{0}{\overset{x}{\int }}g\text{'}\left(t\right)dt=\underset{0}{\overset{x}{\int }}\left(3t^2+5{\text{e}}^t\right)dt$           (A1)

$g\left(x\right)-4=x^3+5{\text{e}}^x-5{\text{e}}^0$           (A1)(A1)

Note: Award A1 for $g\left(x\right)-4$ and A1 for $x^3+5{\text{e}}^x-5{\text{e}}^0$.

$g\left(x\right)=x^3+5{\text{e}}^x-1$           A1

 

[5 marks]

While many students were successful in solving this question, some did not consider the constant of integration or struggled to integrate the exponential term. A few students lost the final mark for stopping at $C=-1$ and not giving the formula for $g\left(x\right)$.