This TSA checks definite integrals with substitution and integration by parts, Riemann sums, modulus and signed area, area between curves, volumes of revolution about the axes, and numerical evaluation of definite integrals using a GC.
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Definite Integrals with Limits and Techniques
1. When using substitution in a definite integral, the integrand, the differential and the limits can be converted to the new variable across different steps.
Tested fundamental: Presenting substitution in a definite integral consistently.
Explanation: Once the definite integral is rewritten using the new variable, the integrand, the differential and the limits must all be consistently expressed in that new variable.
Common wrong choice: Choosing True usually means the student thinks it is acceptable to mix new-variable expressions with old \(x\)-limits.
2. For \[ \int_0^1 \frac{x}{\sqrt{x+1}}\,dx, \] use the substitution \[ u=x+1. \] Which expression is the correctly transformed definite integral?
Tested fundamental: Converting the integrand, differential and limits under substitution.
Explanation: Since \(u=x+1\), we have \(x=u-1\) and \(dx=du\). When \(x=0\), \(u=1\). When \(x=1\), \(u=2\). Hence \(\int_0^1 \frac{x}{\sqrt{x+1}}\,dx=\int_1^2 \frac{u-1}{\sqrt u}\,du.\)
Common wrong choice: A changes the integrand and differential but leaves the old \(x\)-limits unchanged.
3. Which formula correctly shows integration by parts for a definite integral?
Tested fundamental: Applying integration by parts with limits.
Explanation: For definite integrals, the boundary term must also be evaluated between the limits: \(\int_a^b u\,dv=[uv]_a^b-\int_a^b v\,du.\)
Common wrong choice: B is the indefinite version and does not show the boundary evaluation.
Riemann Sums and Exact Area
4. Which expression is the right Riemann sum for \(f(x)\) on \(0\le x\le1\) using \(n\) equal rectangles?
Tested fundamental: Forming right Riemann sums on \([0,1]\).
Explanation: On \([0,1]\), the width is \(\frac1n\). The right endpoints are \(\frac1n,\frac2n,\ldots,\frac nn\). So the right Riemann sum is \(\frac1n\sum_{r=1}^{n}f\left(\frac{r}{n}\right)\).
Common wrong choice: D uses left endpoints, not right endpoints.
5. If \(f\) is increasing on \([0,1]\), which statement is correct?
Tested fundamental: Overestimation and underestimation using left and right rectangles.
Explanation: For an increasing function, left-endpoint rectangles lie below the curve, while right-endpoint rectangles lie above the curve.
Common wrong choice: A reverses the behaviour of left and right rectangles for an increasing function.
6. For a continuous function \(f\), \[ \lim_{n\to\infty} A_n=\int_a^b f(x)\,dx, \] where \(A_n\) is the area approximation using \(n\) rectangles.
Tested fundamental: Connecting Riemann sums with exact definite integrals.
Explanation: As the number of rectangles increases, the width of each rectangle tends to \(0\). The Riemann sum then approaches the exact definite integral.
Common wrong choice: Choosing False usually means the student sees Riemann sums only as approximations and misses the limiting process.
7. Which definite integral is represented by \[ \lim_{n\to\infty} \frac{2}{n} \sum_{r=1}^{n} \left(1+\frac{2r}{n}\right)^3? \]
Tested fundamental: Translating a limit of sum into a definite integral.
Explanation: The width is \(\frac{2}{n}\), so the interval length is \(2\). The right endpoints are \(1+\frac{2r}{n}\), which run from \(1\) to \(3\). Hence the limit represents \(\int_1^3 x^3\,dx\).
Common wrong choice: C treats the width \(\frac{2}{n}\) as part of the integrand instead of using it to identify the interval length.
8. Evaluate \[ \lim_{n\to\infty} \frac1n\sum_{r=1}^{n}\left(\frac{r}{n}\right)^2. \]
Tested fundamental: Evaluating a limit of sum as a definite integral.
Explanation: The expression represents the right Riemann sum for \(f(x)=x^2\) on \(0\le x\le1\). Hence \(\lim_{n\to\infty}\frac1n\sum_{r=1}^{n}\left(\frac{r}{n}\right)^2=\int_0^1x^2\,dx=\frac13.\)
Common wrong choice: B confuses the result with \(\int_0^1 x\,dx\).
Modulus, Signed Area and Area Between Curves
9. Which expression correctly rewrites \[ \int_{-2}^{3}|x-1|\,dx? \]
Tested fundamental: Splitting a modulus integral at the point where the expression changes sign.
Explanation: The expression \(x-1\) is negative when \(x<1\) and positive when \(x>1\). Hence \(|x-1|=1-x\) for \(x<1\), and \(|x-1|=x-1\) for \(x>1\).
Common wrong choice: B ignores the modulus and treats the signed integral as the required positive area.
10. For any function \(f\), \[ \int_a^b |f(x)|\,dx = \left|\int_a^b f(x)\,dx\right|. \]
Tested fundamental: Distinguishing signed integral from total area.
Explanation: This is not true in general. If \(f(x)\) changes sign, positive and negative signed areas may cancel in \(\int_a^b f(x)\,dx\), but \(\int_a^b |f(x)|\,dx\) counts all areas as positive.
Common wrong choice: Choosing True usually means the student has confused signed area with total area.
11. Find the exact area bounded by \[ y=|x^2-2x|, \] the \(x\)-axis, and the lines \[ x=0,\qquad x=3. \]
Answer format: Give your answer as an exact fraction. Use an improper fraction if applicable.
Answer: \(\displaystyle \frac83\)
Tested fundamental: Finding area under a modulus function by splitting at sign changes.
Explanation: First consider \(x^2-2x=x(x-2)\). This is negative for \(0<x<2\), and positive for \(2<x<3\). Hence \(|x^2-2x|=2x-x^2\) on \(0\le x\le2\), and \(|x^2-2x|=x^2-2x\) on \(2\le x\le3\).
So the required area is \(\int_0^2(2x-x^2)\,dx+\int_2^3(x^2-2x)\,dx=\frac43+\frac43=\frac83.\)
Common error: Integrating \(x^2-2x\) directly gives signed area, not the required total area.
12. Suppose \(f(x)\ge g(x)\) for \(a\le x\le b\). Which expression gives the area between the curves \(y=f(x)\) and \(y=g(x)\) from \(x=a\) to \(x=b\)?
Tested fundamental: Setting up area between two curves.
Explanation: Area between two curves is upper curve minus lower curve. Since \(f(x)\ge g(x)\), the area is \(\int_a^b [f(x)-g(x)]\,dx.\)
Common wrong choice: B reverses upper and lower curve, giving a negative value.
13. The function \(f\) is defined by \[ f(x)= \begin{cases} x^2, & 0\le x<1,\\ 2-x, & 1\le x<2, \end{cases} \] and \[ f(x)=f(x+2). \] Which expression gives the area bounded by \(y=f(x)\) and the \(x\)-axis for \[ -1\le x\le3? \]
Tested fundamental: Splitting area under a periodic piecewise function using the original pieces.
Explanation: The function has period \(2\). For \(-1\le x<0\), the corresponding part in the original period is \(1\le x<2\), so we use \(\int_1^2(2-x)\,dx\). For \(0\le x<1\), use \(\int_0^1x^2\,dx\). For \(1\le x<2\), use \(\int_1^2(2-x)\,dx\). For \(2\le x\le3\), the corresponding part in the original period is \(0\le x\le1\), so use \(\int_0^1x^2\,dx\).
Common wrong choice: A gives the same total value here, but it skips the interval-by-interval setup from the required interval \([-1,3]\).
Volumes of Revolution About the Axes
14. The region under \(y=f(x)\), above the \(x\)-axis, from \(x=a\) to \(x=b\), is rotated through \(2\pi\) radians about the \(x\)-axis. Which formula gives the volume generated?
Tested fundamental: Volume of revolution about the \(x\)-axis using discs.
Explanation: Rotation about the \(x\)-axis gives circular discs with radius \(f(x)\). The cross-sectional area is \(\pi[f(x)]^2\). Hence \(V=\pi\int_a^b [f(x)]^2\,dx.\)
Common wrong choice: A gives an area under a curve, not a volume of revolution.
15. The same region under \(y=f(x)\), from \(x=a\) to \(x=b\), is rotated through \(\pi\) radians about the \(x\)-axis. Which formula gives the volume generated?
Tested fundamental: Adjusting volume of revolution for a half-turn.
Explanation: A rotation of \(\pi\) radians is a half-turn. It generates half the volume formed by a full \(2\pi\)-radian rotation. Hence the coefficient is \(\frac{\pi}{2}\).
Common wrong choice: B uses the full-turn formula instead of the half-turn formula.
16. The region bounded by \(x=f(y)\), the \(y\)-axis, \(y=a\), and \(y=b\) is rotated through \(2\pi\) radians about the \(y\)-axis. Which formula gives the volume generated?
Tested fundamental: Volume of revolution about the \(y\)-axis using \(x=f(y)\).
Explanation: Rotation about the \(y\)-axis gives discs with radius \(x=f(y)\). Therefore \(V=\pi\int_a^b [f(y)]^2\,dy.\)
Common wrong choice: A uses \(x\)-limits and \(f(x)\), which is not the correct setup for this description.
17. The region between \[ y=4 \] and \[ y=x^2, \] for \[ 0\le x\le2, \] is rotated about the \(x\)-axis. Which setup gives the volume generated?
Tested fundamental: Finding a volume by subtracting the volume under a curve from a cylinder.
Explanation: The region can be viewed as the rectangle under \(y=4\) from \(x=0\) to \(x=2\), excluding the region under \(y=x^2\). When the rectangle is rotated about the \(x\)-axis, it forms a cylinder with radius \(4\) and height \(2\). Its volume is \(\pi(4)^2(2)=32\pi\).
The region under \(y=x^2\) generates volume \(\pi\int_0^2(x^2)^2\,dx=\pi\int_0^2x^4\,dx\). Therefore the required setup is \(32\pi-\pi\int_0^2x^4\,dx.\)
Common wrong choices: A uses \(16\pi\), which comes from confusing the radius and height of the cylinder. C uses \(x^2\) instead of \((x^2)^2=x^4\).
18. Suppose \(f(y)\ge g(y)\ge0\) for \(a\le y\le b\). The region between \(x=f(y)\) and \(x=g(y)\) is rotated about the \(y\)-axis. Which formula gives the volume generated?
Tested fundamental: Washer method about the \(y\)-axis.
Explanation: For rotation about the \(y\)-axis, the outer radius is \(f(y)\), and the inner radius is \(g(y)\). Hence \(V=\pi\int_a^b\left([f(y)]^2-[g(y)]^2\right)\,dy.\)
Common wrong choice: A squares the difference between the radii instead of subtracting the squares of the radii.
19. A region is bounded by \[ y=x^2 \] and \[ y=4. \] It is rotated through \(\pi\) radians about the \(y\)-axis. Which setup gives the volume generated?
Tested fundamental: Recognising symmetry in volume of revolution.
Explanation: The region is symmetric about the \(y\)-axis. Although the rotation is only through \(\pi\) radians, the half-turn already sweeps out the full circular cross-section because the original horizontal strip stretches from \(x=-\sqrt y\) to \(x=\sqrt y\).
At height \(y\), the radius is \(\sqrt y\). So the cross-sectional area is \(\pi(\sqrt y)^2=\pi y\). Hence \(V=\pi\int_0^4 y\,dy.\)
Common wrong choice: A incorrectly halves the volume even though the original region is already symmetric about the axis of rotation.
Evaluation of Definite Integrals Using GC
20. Two curves enclose a finite region. Their intersection points cannot be found exactly, and a numerical answer for the area is acceptable. Which process correctly finds the area?
Tested fundamental: Using GC/software appropriately for non-exact intersections in an area problem.
Explanation: When the intersection points are non-exact, GC/software can be used to find the required limits. The area must still be set up as \(\int(\text{upper curve}-\text{lower curve})\,dx\) over the correct interval.
Common wrong choice: A changes the limits to convenient values, which changes the required region.
21. The region under \[ y=e^{-x^2}, \] above the \(x\)-axis, from \(x=0\) to \(x=1\), is rotated through \(2\pi\) radians about the \(x\)-axis. Find the volume generated, correct to \(2\) decimal places.
Answer format: Give your answer correct to 2 decimal places.
Answer: \(1.88\)
Tested fundamental: Setting up and evaluating a volume of revolution using GC/software.
Explanation: For rotation about the \(x\)-axis, \(V=\pi\int_0^1\left(e^{-x^2}\right)^2dx\). So \(V=\pi\int_0^1 e^{-2x^2}\,dx\).
This definite integral does not have a simple elementary exact form, so GC/software is appropriate. Evaluating gives \(V\approx1.88\).
Common error: Forgetting to square \(e^{-x^2}\) before integrating.
22. Using the substitution \[ u=x^2+1, \] evaluate \[ \int_0^1 2x e^{(x^2+1)^2}\,dx \] correct to \(3\) significant figures.
Answer format: Give your answer to 3 significant figures.
Answer: \(15.0\)
Tested fundamental: Using substitution with changed limits before evaluating numerically.
Explanation: Since \(u=x^2+1\), we have \(du=2x\,dx\). When \(x=0\), \(u=1\). When \(x=1\), \(u=2\). Therefore \(\int_0^1 2x e^{(x^2+1)^2}\,dx=\int_1^2 e^{u^2}\,du.\)
This definite integral does not have a simple elementary exact form, so GC/software is appropriate. Evaluating gives \(\int_1^2 e^{u^2}\,du\approx15.0\) to \(3\) significant figures.
Common error: Leaving the original \(x\)-limits \(0\) and \(1\) after changing the integral to \(u\).
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