A gradient function is given by \(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 10{{\rm{e}}^{2x}} - 5\) . When \(x = 0\) , \(y = 8\) . Find the value of y when \(x = 1\) .

METHOD 1

evidence of anti-differentiation     (M1)

e.g. \(\int {(10{{\rm{e}}^{2x}} - 5){\rm{d}}x} \)

\(y = 5{{\rm{e}}^{2x}} - 5x + C\)     A2A1

Note: Award A2 for \(5{{\rm{e}}^{2x}}\) , A1 for \( - 5x\) . If “C” is omitted, award no further marks.

substituting \((0{\text{, }}8)\)     (M1)

e.g. \(8 = 5 + C\)

\(C = 3\)  \((y = 5{{\rm{e}}^{2x}} - 5x + 3)\)     (A1)

substituting \(x = 1\)     (M1)

\(y = 34.9\)  \((5{{\rm{e}}^2} - 2)\)     A1     N4

METHOD 2

evidence of definite integral function expression     (M2)

e.g. \(\int_a^x {f'(t){\rm{d}}t = } f(x) - f(a)\) , \(\int_0^x {(10{{\rm{e}}^{2x}} - 5)} \)

initial condition in definite integral function expression     (A2)

e.g. \(\int_0^x {(10{{\rm{e}}^{2t}} - 5)} {\rm{d}}t = y - 8\) , \(\int_0^x {(10{{\rm{e}}^{2x}} - 5)} {\rm{d}}x + 8\)

correct definite integral expression for y when \(x = 1\)     (A2)

e.g. \(\int_0^1 {(10{{\rm{e}}^{2x}} - 5){\rm{d}}x + 8} \)

\(y = 34.9\)  \((5{{\rm{e}}^2} - 2)\)     A1     N4

[8 marks]

Although a pleasing number of candidates recognized the requirement of integration, many did not correctly apply the reverse of the chain rule to integration. While some candidates did not write the constant of integration, many did, earning additional follow-through marks even with an incorrect integral. Weaker candidates sometimes substituted \(x = 1\) into \(\frac{{{\rm{d}}y}}{{{\rm{d}}x}}\) or attempted some work with a tangent line equation, earning no marks.