Integration and the fundamental theorem of calculus – university Year 1 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Use integration by parts, choosing \(u=x\) (it becomes simpler when differentiated) and \(v'=\cos x\), so \(u'=1\) and \(v=\sin x\) (1). Then \(\int x\cos x\,dx=x\sin x-\int\sin x\,dx=\) \(x\sin x+\cos x+C\) (1). Differentiating gives back \(x\cos x\).
   2. Let \(u=1+x^2\), so \(du=2x\,dx\); the limits become \(1\) and \(2\). The integral is \(\tfrac12\int_1^2\dfrac{du}u\) (1) \(=\tfrac12\left[\ln u\right]_1^2=\) \(\tfrac12\ln2\) (1), about \(0.347\).

   In (a), choosing \(u=\cos x\) instead makes the new integral \(\int\tfrac{x^2}2\sin x\,dx\) harder, so the choice of \(u\) matters.

2. Answer to Question 2 (4 marks)

   First split into partial fractions: \(1=A(x+1)+B(x-1)\); \(x=1\) gives \(A=\tfrac12\) and \(x=-1\) gives \(B=-\tfrac12\), so \(\dfrac1{(x-1)(x+1)}=\dfrac12\left(\dfrac1{x-1}-\dfrac1{x+1}\right)\) (2). Integrating each term and combining the logarithms gives \(\tfrac12\left[\ln\dfrac{x-1}{x+1}\right]_2^3\) (1). Substituting the limits, \(\tfrac12\left(\ln\dfrac24-\ln\dfrac13\right)=\tfrac12\left(\ln\dfrac12-\ln\dfrac13\right)=\tfrac12\ln\dfrac32\) (1), as a single logarithm. Since \(\frac32>1\), the answer is positive (about \(0.203\)), as expected for a positive integrand on \([2,3]\).

   A common error is to integrate \(\tfrac1{(x-1)(x+1)}\) directly as \(\ln|(x-1)(x+1)|\); partial fractions are needed first. The integrand is continuous on \([2,3]\), so the integral is proper.

3. Answer to Question 3 (4 marks)

   1. The area under \(y=x^2\) from \(0\) to \(2\) is \(\int_0^2x^2dx=\left[\dfrac{x^3}3\right]_0^2=\dfrac83\) square units (1).
   2. Rotating about the \(x\)-axis, each slice is a disc of radius \(y=x^2\), so \(V=\pi\int_0^2y^2dx=\pi\int_0^2x^4dx\) (1) \(=\pi\left[\dfrac{x^5}5\right]_0^2=\dfrac{32\pi}5\approx20.1\) cubic units (1).
   3. The mean value is \(\dfrac1{b-a}\int_a^by\,dx=\dfrac12\cdot\dfrac83=\dfrac43\) (1). A rectangle of height \(\frac43\) on \([0,2]\) has the same area as \(R\).

   In (b), a common error is to use \(\pi\int_0^2x^2\,dx\); squaring the radius is essential because each disc has area \(\pi y^2\). In (c), dividing by the interval length 2 is the key step.

4. Answer to Question 4 (4 marks)

   1. Replace the infinite limit by \(b\) and take a limit: \(\int_1^bx^{-3/2}dx=\left[-2x^{-1/2}\right]_1^b=2-\dfrac2{\sqrt b}\) (1). As \(b\to\infty\), \(\dfrac2{\sqrt b}\to0\), so the integral converges to 2 (1).
   2. Similarly \(\int_1^bx^{-1/2}dx=\left[2\sqrt x\right]_1^b=2\sqrt b-2\) (1). This tends to \(\infty\) as \(b\to\infty\), so the integral diverges (1). In general \(\int_1^\infty x^{-p}dx\) converges only when \(p>1\); here \(p=\frac12\), so the area under the curve is infinite even though \(\frac1{\sqrt x}\to0\).

5. Answer to Question 5 (4 marks)

   1. Let \(F(u)=\int_1^u\ln t\,dt\), so \(F'(u)=\ln u\) by the fundamental theorem. By the chain rule, \(\dfrac d{dx}F(x^3)=\ln(x^3)\cdot3x^2\) (1) \(=3\ln x\cdot3x^2=9x^2\ln x\) (1).
   2. The width of each strip is \(\frac14\), and the right endpoints are \(\frac14,\frac12,\frac34,1\). The sum is \(\frac14\left(\frac1{16}+\frac14+\frac9{16}+1\right)\) (1) \(=\frac14\cdot\frac{30}{16}=\frac{15}{32}\approx0.469\) (1). It overestimates the exact value \(\frac13\), because \(x^2\) is increasing.

   In (a), a common error is to give only \(\ln(x^3)\), forgetting the chain rule factor \(3x^2\) from the upper limit. In (b), the error of the estimate is \(\tfrac{15}{32}-\tfrac13=\tfrac{13}{96}\approx0.135\).

Total : 20 marks