## Question 3(i):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int x^2 e^{2x}\, dx = \frac{1}{2}x^2e^{2x} - \int xe^{2x}\, dx$ | M1 | Attempts integration by parts once to achieve $Px^2e^{2x} - Q\int xe^{2x}\,dx$, $P,Q>0$ |
| $= \frac{1}{2}x^2e^{2x} - \left\{\frac{1}{2}xe^{2x} - \int \frac{e^{2x}}{2}\,dx\right\}$ | M1 | Attempts to integrate $xe^{2x}$ to achieve $\lambda xe^{2x} \pm \mu\int e^{2x}\,dx$, $\lambda>0$ |
| $= \frac{1}{2}x^2e^{2x} - \frac{1}{2}xe^{2x} + \frac{e^{2x}}{4}$ | A1 | May be unsimplified; isw once correct answer seen. Watch for $\frac{1}{2}x^2e^{2x} - \left(\frac{1}{2}xe^{2x} - \frac{e^{2x}}{4}\right)(+c)$ which scores all 3 marks |
| $\int_0^4 x^2e^{2x}\,dx = \left[\frac{1}{2}x^2e^{2x} - \frac{1}{2}xe^{2x} + \frac{e^{2x}}{4}\right]_0^4 = 8e^8 - 2e^8 + \frac{e^8}{4} - \frac{1}{4}$ | M1 | Attempts to substitute **both** limits into expression of form $Ax^2e^{2x} \pm Bxe^{2x} \pm Ce^{2x}$, $A>0$; subtracts and simplifies to $\alpha e^8 + \beta$, $\alpha,\beta\neq 0$ |
| $= \frac{25e^8}{4} - \frac{1}{4}$ | A1 | Or exact equivalent e.g. $\frac{1}{4}(25e^8-1)$, $\frac{25}{4}\left(e^8 - \frac{1}{25}\right)$. Do **not** condone "$+c$" |

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## Question 3(ii):

| Working/Answer | Mark | Guidance |
|---|---|---|
| **Sub 1:** $u=2x-1 \Rightarrow \frac{du}{dx}=2$ or $du=2\,dx$ **OR Sub 2:** $u=(2x-1)^2 \Rightarrow \frac{du}{dx}=4(2x-1)$ **OR Sub 3:** $u=2x \Rightarrow \frac{du}{dx}=2$ | B1 | Uses appropriate substitution and differentiates **correctly** |
| **Sub 1:** $\int \frac{4x}{(2x-1)^2}\,dx = \int \frac{2u+2}{u^2}\cdot\frac{1}{2}\,du$ **OR Sub 2:** $\int \frac{4x}{(2x-1)^2}\,dx = \int \frac{2\sqrt{u}+2}{u}\cdot\frac{1}{4\sqrt{u}}\,du$ **OR Sub 3:** $\int \frac{4x}{(2x-1)^2}\,dx = \int \frac{2u}{(u-1)^2}\cdot\frac{1}{2}\,du$ | M1 A1 | M1: Valid attempt to change to integrand in $u$ of correct form e.g. $k\int\frac{au+b}{u^2}\,du$ or $k\int\frac{a\sqrt{u}+b}{u}\cdot\frac{1}{\sqrt{u}}\,du$ or $k\int\frac{u}{(u-1)^2}\,du$. A1: Correct integrand in $u$ |
| **Sub 1:** $\int\left(\frac{1}{u}+\frac{1}{u^2}\right)du = \ln u - \frac{1}{u}$ **OR Sub 2:** $\frac{1}{2}\int\left(\frac{1}{u}+\frac{1}{u^{3/2}}\right)du = \frac{1}{2}\ln u - \frac{1}{\sqrt{u}}$ **OR Sub 3:** $\int\frac{u}{(u-1)^2}\,du = -u(u-1)^{-1} + \ln(u-1)$ | dM1 A1 | dM1: Correct form of integration for their substitution (depends on previous M). A1: Fully correct integration |
| **Sub 1:** limits $u=5$ to $u=20$: $\left(\ln 20 - \frac{1}{20}\right) - \left(\ln 5 - \frac{1}{5}\right)$ **OR Sub 2:** limits $u=25$ to $u=400$ **OR Sub 3:** limits $u=6$ to $u=21$ | M1 | Attempts to use correct limits for their substitution; both limits substituted and subtracted. Must have attempted integration. Substitution must involve a transformation so $u=x$ not acceptable. Alternatively reverses substitution and uses $x$ limits 3 and $\frac{21}{2}$ |
| $= \frac{3}{20} + \ln 4$ | A1 | Or e.g. $\ln 4 + 0.15$; apply isw if necessary |