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Test solutions College : Applications of Integration — Zyro the alien explorer of Planète Maths

Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

1 Area between a parabola and a line / 4 pts

(a) \(x^2=2x\) gives \(x=0\) or \(x=2\), so the points are \((0,0)\) and \((2,4)\). (1 pt)

(b) At \(x=1\): \(2>1\), so \(y=2x\) is on top. (1 pt)

(c) \(A=\int_0^2(2x-x^2)\,dx=\Big[x^2-\tfrac{x^3}{3}\Big]_0^2=4-\tfrac83=\tfrac43\). (2 pts)

2 Volume with disks / 4 pts

(a) \(V=\pi\int_0^1(e^x)^2\,dx=\pi\int_0^1e^{2x}\,dx\). (2 pts)

(b) \(V=\pi\Big[\tfrac{e^{2x}}{2}\Big]_0^1=\tfrac{\pi(e^2-1)}{2}\). (1 pt)

(c) \(V\approx 10.04\) cubic units. (1 pt)

3 Volume with shells / 4 pts

Radius \(x\), height \(3x-x^2\). (1 pt) \(V=2\pi\int_0^3x(3x-x^2)\,dx\). (1 pt)

\(=2\pi\Big[x^3-\tfrac{x^4}{4}\Big]_0^3=2\pi\Big(27-\tfrac{81}{4}\Big)\). (1 pt)

\(=\tfrac{27\pi}{2}\approx 42.41\) cubic units. (1 pt)

4 Arc length / 3 pts

\(y'=x^2-\tfrac{1}{4x^2}\), so \(1+(y')^2=\Big(x^2+\tfrac{1}{4x^2}\Big)^2\). (1 pt)

\(L=\int_1^2\Big(x^2+\tfrac{1}{4x^2}\Big)dx=\Big[\tfrac{x^3}{3}-\tfrac{1}{4x}\Big]_1^2\). (1 pt)

\(=\Big(\tfrac83-\tfrac18\Big)-\Big(\tfrac13-\tfrac14\Big)=\tfrac{59}{24}\approx 2.46\). (1 pt)

5 Average value and work / 3 pts

(a) \(\tfrac1\pi\int_0^\pi\sin x\,dx=\tfrac{2}{\pi}\approx0.637\). (1 pt)

(b) \(k=6/0.15=40\) N/m. (1 pt) \(W=\tfrac12(40)(0.3)^2=1.8\) J. (1 pt)

6 Separable equation / 2 pts

Separate: \(\dfrac{dy}{y}=2x\,dx\), so \(\ln|y|=x^2+C\). (1 pt)

\(y(0)=1\) gives \(C=0\), so \(y=e^{x^2}\). (1 pt)

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