Math 152 Midterm 1 Review Notes

Math 152 Midterm 1 Review Sheet (Spring 2025)

Disclaimers

  • The review problems are designed to help you study for the exam.
  • Everything discussed in lecture, recitation, or textbook sections covered is fair game for the exam.
  • This includes material standard to Math 151 or pre-calculus.
  • Individual lecturer's guidance takes precedence over this sheet if there are contradictions.

Exam Rules

  • The exam is closed book.
  • No calculators or devices are allowed.
  • No formula sheet is provided.

Things to have memorized

  • It's expected that you know basic facts such as sin(π/6)=1/2sin(π/6) = 1/2 and arctan(1)=π/4arctan(1) = π/4.
Basic Antiderivatives
  1. xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C, (n1)(n \neq -1)
  2. 1xdx=lnx+C\int \frac{1}{x} dx = ln |x| + C
  3. exdx=ex+C\int e^x dx = e^x + C
  4. axdx=axln(a)+C\int a^x dx = \frac{a^x}{ln(a)} + C, (a > 0, a \neq 1)
  5. cos(x)dx=sin(x)+C\int cos(x) dx = sin(x) + C
  6. sin(x)dx=cos(x)+C\int sin(x) dx = -cos(x) + C
  7. sec(x)dx=lnsec(x)+tan(x)+C\int sec(x) dx = ln |sec(x) + tan(x)| + C
  8. csc(x)dx=lncsc(x)+cot(x)+C\int csc(x) dx = - ln | csc(x) + cot(x)| + C
  9. tan(x)dx=lnsec(x)+C\int tan(x) dx = ln |sec(x)| + C
  10. cot(x)dx=lnsin(x)+C\int cot(x) dx = ln |sin(x)| + C
  11. sec2(x)dx=tan(x)+C\int sec^2(x) dx = tan(x) + C
  12. csc2(x)dx=cot(x)+C\int csc^2(x) dx = - cot(x) + C
  13. sec(x)tan(x)dx=sec(x)+C\int sec(x) tan(x) dx = sec(x) + C
  14. csc(x)cot(x)dx=csc(x)+C\int csc(x) cot(x) dx = - csc(x) + C
  15. 1a2x2dx=sin1(xa)+C\int \frac{1}{\sqrt{a^2 - x^2}} dx = sin^{-1} (\frac{x}{a}) + C
  16. 1a2+x2dx=1atan1(xa)+C\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} tan^{-1} (\frac{x}{a}) + C
  17. 1xx2a2=1asec1xa+C\int \frac{1}{x\sqrt{x^2 - a^2}} = \frac{1}{a} sec^{-1} |\frac{x}{a}| + C
Derivatives
  • Derivatives of all six trigonometric functions.
  • Derivatives of arcsin(x), arcsec(x) and arctan(x).
Disc/Washer Method
  • abπ(R(x)2r(x)2)dx\int_{a}^{b} π (R(x)^2 - r(x)^2) dx
  • cdπ(R(y)2r(y)2)dy\int_{c}^{d} π (R(y)^2 - r(y)^2) dy
Shell Method
  • ab2π(shell radius)(shell height)dx\int_{a}^{b} 2π (shell \ radius)(shell \ height) dx
  • cd2π(shell radius)(shell height)dy\int_{c}^{d} 2π (shell \ radius)(shell \ height) dy
Arc Length
  • Arc length of a curve y=f(x)y = f(x) between x=ax = a and x=bx = b: L=ab1+(f(x))2dxL = \int_{a}^{b} \sqrt{1 + (f'(x))^2} dx
Surface Area
  • Area of the surface obtained by revolving the curve y=f(x)y = f(x) about the x-axis between x=ax = a and x=bx = b: S=2πabf(x)1+(f(x))2dxS = 2π \int_{a}^{b} f(x)\sqrt{1 + (f'(x))^2} dx
Standard Trigonometric Identities
  • sin2(x)+cos2(x)=1sin^2(x) + cos^2(x) = 1
  • 1+tan2(x)=sec2(x)1 + tan^2(x) = sec^2(x)
  • cot2(x)+1=csc2(x)cot^2(x) + 1 = csc^2(x)
  • cos(2x)=cos2(x)sin2(x)cos(2x) = cos^2(x) - sin^2(x)
  • sin(2x)=2sin(x)cos(x)sin(2x) = 2 sin(x) cos(x)
  • cos2(x)=1+cos(2x)2cos^2(x) = \frac{1 + cos(2x)}{2}
  • sin2(x)=1cos(2x)2sin^2(x) = \frac{1 - cos(2x)}{2}

Practice Problems

Section 5.3
  • Average Value

    1. Find the average value on the given interval.

      • f(x)=x22f(x) = - \frac{x^2}{2} on [0,3][0, 3]
      • f(x)=sec(3x)tan(3x)f(x) = sec(3x) tan(3x) on [0,π12][0, \frac{π}{12}]
      • f(x)=3100x2f(x) = \frac{3}{\sqrt{100 - x^2}} on [0,5][0, 5]
      • f(x)=cos(5x)sin(5x)f(x) = cos(5x)\sqrt{sin(5x)} on [π30,π10][\frac{π}{30}, \frac{π}{10}]
Sections 5.5, 5.6
  1. Evaluate the following integrals.

    • x3ex4dx\int x^3e^{x^4} dx
    • earcsin(x)1x2dx\int \frac{e^{arcsin(x)}}{\sqrt{1 - x^2}} dx
    • 1x311x2dx\int \frac{1}{x^3\sqrt{1 - \frac{1}{x^2}}} dx
    • e3e6dttln(t)\int_{e^3}^{e^6} \frac{dt}{tln(t)}
    • cos2(4x)dx\int \cos^2(4x) dx
    • sin5(4x)cos(4x)dx\int sin^5(4x) cos(4x) dx
    • 6x2+8x+17dx\int \frac{6}{x^2 + 8x + 17} dx
    • 0π/9sec(3x)tan(3x)(5sec(3x)4)2dx\int_{0}^{π/9} \frac{sec(3x) tan(3x)}{(5 sec(3x) - 4)^2} dx
    • dx8xx2\int \frac{dx}{\sqrt{8x - x^2}}
  2. Find the areas of the regions enclosed by the lines and curves listed below:

    • y=x2y = x^2 and y=x2+4y = -x^2 + 4
    • y=1x+1y = \frac{1}{x+1} and y=2x+1y = -2x + 1
    • xy2=0x - y^2 = 0 and x+2y2=3x + 2y^2 = 3
    • y=43+x2y = \frac{4}{3+x^2} and y=1y = 1
  3. For the regions sketched below, set-up an expression that calculates the area of the region using (i) a dx-integral, and (ii) a dy-integral. You do not need to evaluate the integrals.

    • The region to the right of the curve y=x2+1y = x^2 + 1, above the line y=1y = 1 and to the left of the line y=92xy = 9 - 2x.
    • The region to the right of the y-axis, below the curve x=9y2x = 9 - y^2 and above the line 2x+9y=182x + 9y = 18.
Sections 6.1, 6.2
  1. Find the volumes of solids generated by revolving the regions bounded by the lines and curves below around the x-axis.

    • y=x3y = x^3, y=0y = 0 and x=2x = 2
    • y=2xy = 2\sqrt{x}, y=2y = 2 and x=0x = 0
    • x=2yy2x = 2y - y^2, x=0x = 0
    • y=xy = \sqrt{x}, y=0y = 0, y=2xy = 2 - x
  2. Find the volumes of solids generated by revolving the regions bounded by the lines and curves below around the y-axis.

    • y=xy = x, y=x/2y = -x/2, x=2x = 2
    • y=2x1y = 2x - 1, y=xy = \sqrt{x}, x=0x = 0
    • x=y3/2x = y^{3/2}, x=0x = 0, y=2y = 2
    • x=2y+1x = \sqrt{2y+1}, x=0x = 0, y=0y = 0, y=3y = 3
  3. Find the volume obtained by revolving each region described below about the indicated axis.

    • The region bounded by the curves y=x2+1y = x^2 + 1 and y=x+3y = x + 3 revolved about the line y=1y = 1.
    • The region to the right of the y-axis bounded by y=x2y = x^2 and y=2xy = 2 - x revolved about the line x=3x = -3.
    • The region bounded by y=exy = e^{-x}, y=0y = 0, x=0x = 0 and x=1x = 1 revolved about the line y=2y = 2.
    • The region bounded by the lines 2y=x+42y = x + 4, y=xy = x and x=0x = 0 revolved about the line x=4x = 4.
  4. Consider the region R bounded by the curve y=ln(x1)y = ln(x - 1), the line x=2x = 2 and the line y=3y = 3.

    • Set-up integrals computing the volume of the solid obtained by revolving R about each axis given below using (i) the disk/washer method, and (ii) the shell method. Do not evaluate the integrals.
      • y=3y = 3
      • y-axis
      • y=4y = -4
      • x=1x = 1
  5. The triangular region bounded by the line y=62xy = 6-2x, the x-axis and the y-axis is the base of a solid. Find the volume of the solid if

    • the cross-sections perpendicular to the x-axis are semi-circles with diameter in the base.
    • the cross-sections perpendicular to the y-axis are squares.
Sections 6.3, 6.4
  1. Find the arc lengths of the following curves on the given intervals:

    • f(x)=x3/2f(x) = x^{3/2} on 0x40 \le x \le 4
    • f(y)=y3/23y1/2f(y) = \frac{y^{3/2}}{3} - y^{1/2} on 1y91 \le y \le 9
    • f(x)=x44+18x2f(x) = \frac{x^4}{4} + \frac{1}{8x^2} on 1x21 \le x \le 2
    • f(x)=ex+ex2f(x) = \frac{e^x+e^{-x}}{2} on 1x1-1 \le x \le 1
    • f(x)=13ln(cos(3x))f(x) = \frac{1}{3} ln(|cos(3x)|) on π4xπ3\frac{π}{4} \leqslant x \leqslant \frac{π}{3}
    • f(x)=36x2f(x) = \sqrt{36 - x^2} for 0x30 \leqslant x \leqslant 3.
  2. Find the areas of the surface generated when the given curves are revolved around the given axis.

    • y=3x+4y = 3x + 4 for 0x60 \le x \le 6 around the x-axis.
    • y=8xy = 8\sqrt{x} for 9x209 \le x \le 20 around the x-axis.
    • y=x24y = \frac{x^2}{4} for 2x42 \le x \le 4 around the y-axis.
    • x=16yy2x = \sqrt{16y - y^2} for 0y80 \leqslant y \leqslant 8 revolved about the y-axis.
  3. Let C be the curve y=ln(2x)y = ln(2x) on the interval 2x72 \le x \le 7. Set up (but do not evaluate) an integral to find the area of the surface generated when C is revolved about the specified axis.

    • Axis: y=2y = -2 (set up as a dx integral)
    • Axis: y=2y = -2 (set up as a dy integral)
    • Axis: x=10x = 10 (set up as a dx integral)
    • Axis: x=10x = 10 (set up as a dy integral)
Section 8.2
  1. Evaluate the following integrals:

    • x2cos(x)dx\int x^2 cos(x) dx
    • xsec2(x)dx\int x sec^2(x) dx
    • lnxx2dx\int \frac{ln x}{x^2} dx
    • xsin(2x)dx\int x sin(2x) dx
    • arcsin(x)dx\int arcsin(x) dx
    • e2xcos(x)dx\int e^{2x} cos(x) dx
    • x2exdx\int x^2e^{-x} dx
    • xx1dx\int \frac{x}{\sqrt{x - 1}} dx
    • xln(x)dx\int \sqrt{x} ln(x) dx
More Practice Problems
  1. Evaluate the following integrals.

    • 1(2x+1)4x+4x2dx\int \frac{1}{(2x + 1)\sqrt{4x + 4x^2}} dx
    • 2x+7x+1dx\int \frac{2x + 7}{\sqrt{x + 1}} dx
    • 1ex+exdx\int \frac{1}{e^x + e^{-x}} dx
    • 1x2+6xdx\int \frac{1}{\sqrt{-x^2 + 6x}} dx
    • 1e2x1dx\int \frac{1}{\sqrt{e^{2x} - 1}} dx
    • 4282xx2dx\int_{-4}^{2} \sqrt{8 - 2x - x^2} dx
    • xex(x+1)2dx\int \frac{xe^x}{(x + 1)^2} dx
    • arctan(x)dx\int arctan(x) dx
    • e8xsin(3x)dx\int e^{8x} sin(3x) dx
    • e5x+3dx\int e^{\sqrt{5x+3}} dx
    • x2arctan(x)dx\int x^2 arctan(x) dx
  2. Find the average value of f(x)=x7+x2+sinx+2xx2+1f(x) = x^7 + x^2 + sin x + |2x|\sqrt{x^2 + 1} on the interval [3,3][-3, 3].

  3. Let F(x)F(x) be an anti-derivative of f(x)=sin(x)xf(x) = \frac{sin(x)}{x} for x > 0. Express 13sin(2x)xdx\int_{1}^{3} \frac{sin(2x)}{x} dx in terms of F(x)F(x).

  4. Find the area of the ‘triangular’ region in the first quadrant that is bounded above by the curve y=ex/2y = e^{x/2}, below by y=ex/2y = e^{-x/2} and on the right by x=2ln(2)x = 2 ln(2).

  5. Consider the region R inside the circle x2+y2=25x^2 + y^2 = 25 and above the horizontal line y=3y = 3.

    • Set up an x-integral that calculates the area of the region R. Do not evaluate the integral.
    • Set up a y-integral that calculates the area of the region R. Do not evaluate the integral.
  6. Let R be the region bounded by x=yy3x = y - y^3 and the y-axis. Find the volume of the solid generated by revolving R around the line y=1y = 1.

  7. Let R be the region bounded by y=2xx2y = 2x - x^2 and y=xy = x. Find the volume of the solid generated by revolving R around the line x=1x = 1.

  8. The region R is bounded by y=ex/2y = e^{x/2}, y=7y = 7 and x=2x = 2. We create a solid of revolution by revolving R about the line x=3x = -3.

    • Set up an integral that computes the volume of the solid using the method of cylindrical shells. Do not evaluate the integral.
    • Set up an integral that computes the volume of the solid using the disk/washer method. Do not evaluate the integral.
  9. The region R in the plane bounded by y=xy = \sqrt{x} and y=x/2y = x/2 is the base of a solid. Find the volume of the solid if:

    • The cross-sections perpendicular to the x-axis are squares.
    • The cross-sections perpendicular to the x-axis are isosceles triangles of height 6.
    • The cross-sections perpendicular to the y-axis are circular disks whose diameters traverse R.
  10. A region R is bounded in the plane by the lines: y=x+3y = -x + 3, y=2x6y = 2x - 6 and y=3y = 3. Find the volume of the solid given by rotating R around the line x=5x = 5.

  11. Set up an integral that computes the perimeter of the ellipse x24+y225=1\frac{x^2}{4} + \frac{y^2}{25} = 1.

  12. Find the length of the curve y=0xcos(2t)dty = \int_{0}^{x} cos(2t) dt from x=0x = 0 to x=π4x = \frac{π}{4}.

  13. Let L be the line segment between (1,2)(1, 2) and (4,4)(4, -4). Find the area of the surface generated when L is revolved around the line y=2y = 2.

  14. Let L be the line segment between (1,2)(1, 2) and (4,4)(4, -4). Find the area of the surface generated when L is revolved around the line x=1x = -1.

  15. Set-up integrals to compute the surface area of the cone whose base has radius 4 and has height 5.

  16. Let R be the region under the graph of y=116x2y = \sqrt{\frac{1}{16-x^2}} for 0x20 \leqslant x \leqslant 2. Find the volume of the solid obtained by revolving R about the line x=3x = -3.

  17. Consider the region R between the graphs of y=sec(x)y = sec(x) and y=tan(x)y = tan(x) for 0xπ40 \leqslant x \leqslant \frac{π}{4}.

    • Find the area of the region.
    • Find the volume of the solid obtained by revolving the region about the x-axis.
    • Find the volume of the solid whose base is the region R and whose cross-sections perpendicular to the x-axis are squares.
  18. Find the arc length of the graph of f(x)=45x5/4f(x) = \frac{4}{5}x^{5/4} for 0x90 \leqslant x \leqslant 9.

  19. Use integration by parts to derive the reduction formula for sinnxdx\int sin^n x dx:

    • sinnxdx=1nsinn1xcosx+n1nsinn2xdx\int sin^n x dx = - \frac{1}{n} sin^{n-1} x cos x + \frac{n - 1}{n} \int sin^{n-2} x dx
  20. Use integration by parts to rewrite csc7(3x)dx\int csc^7(3x) dx in terms of csc5(3x)dx\int csc^5(3x) dx.