AP Calc BC Chapter 6 Practice Problems
find the area of the region bounded by the coordinate axes, the line y=3, and the graph of x=y²+2 with respect to y
Chap. 5 Test, 1 determine where the curves intersect to find bounds determine which curve is "on top" plug into area EQ and solve
find the volume of the shape generated by revolving the area between the curve y= 1/x₂ and the x-axis about the x-axis
Chap. 5 Test, 3 determine if/where curves intersect to find bounds determine which curve is "on top" plug into volume EQ and solve
find y': y=x² * log₂(3-2x)
HW #7, 15 logarithmic derivative rule and chain rule
condense: 2ln(x+1) + 1/3 * lnx - ln(cosx)
HW #7, 15 take coefficients and move to power spot then make into 1 log (pos. on top, neg. on bottom)
find the derivative: y = x∧sinx
HW #7, 32 take the ln of both sides, derive implicitly
integrate: ∫[2/x + 3sinx] dx
HW #7, 59 logarithmic integration rule and chain rule
integrate: ∫(t+1)/t dt
HW #7, 65 split up fraction and integrate terms separately with logarithmic integration rules
integrate: ∫e∧2x dx
HW #7, 65 u substitution
integrate: ∫²₀ 3x/1+x² dx
HW #7, 67 (the first) u substitution and logarithmic integration rule
integrate: ∫e∧sinx * cosx dx
HW #7, 67 (the second) u substitution
Sketch the curve y=1/t and shade a region under the curve whose area is a) ln 2 b) -ln0.5 c) 2.
HW #8, 1 write integral of curve to find area beneath determine bounds, keeping in mind that ln1=0 a) bounds: 1-2 b) bounds: 0.5-1 (1 on top gets negative number) c) bounds: 1-e² (lne² becomes 2lne which is just 2)
find dy/dx: y= tan⁻¹(x³)
HW #8, 19 plug into derivative of inverse trig function equation and chain rule
find dy/dx: y=sec⁻¹(x⁵)
HW #8, 20 plug into derivative of inverse trig function equation and chain rule
find the integral: ∫[1/2√1-x² - 3/1+x²] dx
HW #8, 29 split up into separate integrals, then use the derivative of inverse trig function equations
find the limit: lim(x→infinity) [cos(2/x)]∧x²
HW #8, 30 (the first) plug in 0, put into a good indeterminate form, then use L'Hospitals rule (twice)
evaluate the integral: ∫1/√1-4x² dx
HW #8, 31 take out the square from 4x² so that it is (2x)² use u substitution where u = 2x use derivative of inverse trig function equation to integrate
evaluate the integral: ∫sinx/(cos²x+1) dx
HW #8, 36 rearrange denominated so square is on the right u substitution where u=cosx use derivative of inverse trig function equation to integrate
evaluate the integral: ∫√₂ ² 1/x√x²-1 dx
HW #8, 39 use derivative of inverse trig function equation to integrate plug in bounds (top-bottom) set up triangle so that sec⁻¹x=2 and √2 to find angles plug in angles
find the exact value: sec[sin⁻¹(-3/4)]
HW #8, 4 (the second) draw a triangle where sinx=3/4, fill in the third side with Pythagorean theorem, then find secant of that triangle
evaluate the integral: ∫₁ ∧√e 1/x√1-(lnx)² dx
HW #8, 42 u substitution where u=lnx use derivative of inverse trig function equation to integrate plug in bounds (top-bottom) set up triangle so that sec⁻¹x=2 and √2 to find angles plug in angles
find the exact value: sin[2cos⁻¹(3/5)]
HW #8, 5 call stuff in brackets "theta" and set up double sine formula, draw a triangle where cos=3/5 and fill in the third side with Pythagorean theorem, then find sine and cosine of that triangle and plug those into the double angle formula
find the limit: lim(x→0) (e∧x-1)/sinx
HW #8, 7 plug in 0 then use L'Hospitals rule
find y': y=ln(cos(e∧x))
Quiz, 1 logarithmic derivative rule and chain rule
find y': y = e∧³√2=5x²
Quiz, 2 logarithmic derivative rule and chain rule