Past Exam Questions

Show that \(\int_{0}^{\pi} x^2 \sin x \, dx = \pi^2 - 4\).

Use integration by parts to show that

\(\int_{2}^{4} \ln x \, dx = 6 \ln 2 - 2.\)

Find the exact value of \(\int_{0}^{1} xe^{2x} \, dx\).

Find the exact value of \(\int_{1}^{2} x \ln x \, dx\).

By using integration by parts, show that for all \(a > 1\), \(\int_{1}^{a} \frac{\ln x}{x^4} \, dx < \frac{1}{9}\).

Using integration by parts, find the exact value of \(\int_0^2 \arctan\left(\frac{1}{2}x\right) \, dx\).

The equation of a curve is \(y = x^{-\frac{2}{3}} \ln x\) for \(x > 0\).

Show that \(\int_{1}^{8} y \, dx = 18 \ln 2 - 9\).

Find the exact value of \(\int_{0}^{1} (2-x)e^{-2x} \, dx\).

Find the exact value of

\(\int_{1}^{4} x^{\frac{3}{2}} \ln x \, dx.\)

Find \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} x \sec^2 x \, dx\). Give your answer in a simplified exact form.

(i) By differentiating \(\frac{\cos x}{\sin x}\), show that if \(y = \cot x\) then \(\frac{dy}{dx} = -\csc^2 x\).

(ii) Show that \(\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi} x \csc^2 x \, dx = \frac{1}{4}(\pi + \ln 4)\).

The diagram shows the curve \(y = x \cos 2x\), for \(x \geq 0\).

(a) Find the equation of the tangent to the curve at the point where \(x = \frac{1}{2} \pi\).

(b) Find the exact area of the shaded region shown in the diagram, bounded by the curve and the \(x\)-axis.

The diagram shows the curve \(y = x^2 e^{2-x}\) and its maximum point \(M\).

(i) Show that the \(x\)-coordinate of \(M\) is 2.

(ii) Find the exact value of \(\int_0^2 x^2 e^{2-x} \, dx\).

The diagram shows the curve \(y = x \cos \frac{1}{2}x\) for \(0 \leq x \leq \pi\).

(i) Find \(\frac{dy}{dx}\) and show that \(4 \frac{d^2y}{dx^2} + y + 4 \sin \frac{1}{2}x = 0\).

(ii) Find the exact value of the area of the region enclosed by this part of the curve and the x-axis.

The diagram shows the curve \(y = x^2 \ln x\) and its minimum point \(M\).

(i) Find the exact values of the coordinates of \(M\).

(ii) Find the exact value of the area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = e\).

The diagram shows the curve \(y = x^2 e^{-x}\).

(i) Show that the area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = 3\) is equal to \(2 - \frac{17}{e^3}\).

(ii) Find the \(x\)-coordinate of the maximum point \(M\) on the curve.

(iii) Find the \(x\)-coordinate of the point \(P\) at which the tangent to the curve passes through the origin.

The diagram shows the curve \(y = x^3 \ln x\) and its minimum point \(M\).

(i) Find the exact coordinates of \(M\).

(ii) Find the exact area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = 2\).

The diagram shows the curve \(y = \frac{\ln x}{\sqrt{x}}\) and its maximum point \(M\). The curve cuts the \(x\)-axis at the point \(A\).

(i) State the coordinates of \(A\).

(ii) Find the exact value of the \(x\)-coordinate of \(M\).

(iii) Using integration by parts, show that the area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = 4\) is equal to \(8 \ln 2 - 4\).

The diagram shows a sketch of the curve \(y = x^{\frac{1}{2}} \ln x\) and its minimum point \(M\). The curve cuts the \(x\)-axis at the point \((1, 0)\).

(i) Find the exact value of the \(x\)-coordinate of \(M\).

(ii) Use integration by parts to find the area of the shaded region enclosed by the curve, the \(x\)-axis and the line \(x = 4\). Give your answer correct to 2 decimal places.

The diagram shows the curve \(y = x^2 e^{-\frac{1}{2}x}\).

(i) Find the \(x\)-coordinate of \(M\), the maximum point of the curve.

(ii) Find the area of the shaded region enclosed by the curve, the \(x\)-axis and the line \(x = 1\), giving your answer in terms of \(e\).