(a) Find \(\int \frac{1}{\sqrt{3 x+2}} \mathrm{~d} x\). (b) Find, in terms of a, \(\int_{0.5}^{a} \mathrm{e}^{(1-2 x)} \mathrm{d} x\).

Find the exact area of the region enclosed by the curve \(y=\mathrm{e}^{2-4 x}\), the \(x\)-axis, the line \(x=-0.25\) and the line \(x=0.5\).

(a) Differentiate \(y=2xe^{4x}\) with respect to \(x\).

(b) Hence find \(\int xe^{4x}\,dx\).

A curve with equation \(y=\mathrm f(x)\) is such that

\(\displaystyle \frac{\mathrm d^2y}{\mathrm dx^2}=6e^{3x}+4x.\)

The curve has a gradient of \(5\) at the point \(\left(0,\frac53\right)\). Find \(\mathrm f(x)\).

The equation of a curve is given by

\(y=xe^{-2x}.\)

(i) Find \(\frac{dy}{dx}\).

(ii) Find the exact coordinates of the stationary point on the curve \(y=xe^{-2x}\).

(iii) Find, in terms of \(e\), the equation of the tangent to the curve \(y=xe^{-2x}\) at the point \(\left(1,\frac1{e^2}\right)\).

(iv) Using your answer to part (i), find \(\int xe^{-2x}\,dx\).

The curve \(y=f(x)\) passes through the point \(\left(\dfrac12,\dfrac72\right)\) and is such that

\(f'(x)=e^{2x-1}.\)

(i) Find the equation of the curve.

(ii) Find the value of \(x\) for which \(f''(x)=4\), giving your answer in the form \(a+b\ln\sqrt2\), where \(a\) and \(b\) are constants.

It is given that

\(\int_{-k}^{k}\left(15e^{5x}-5e^{-5x}\right)\,dx=6.\)

(i) Show that

\(e^{5k}-e^{-5k}=3.\)

(ii) Hence, using the substitution \(y=e^{5k}\), or otherwise, find the value of \(k\).