- A particle, moving in a straight line, passes through a fixed-point O. Its velocity v ms-1, t seconds after passing through O, is given by v = t + 2 cos(t/3). Find the displacement of the particle
- A particle, moving in a straight line, passes through a fixed-point O. Its velocity v ms-1, t seconds after passing through O, is given by v = 50/(3t + 2)2.a. Find the velocity of the particle as it
- A particle, moving in a straight-line, passes through a fixed-point O. Its velocity v ms-1, t seconds after passing through O, is given by v = 10t – t2.a. Find the velocity of the particle when the
- The velocity-time graph represents the motion of a particle moving in a straight line.a. Find the acceleration during the first 5 seconds.b. Find the length of time for which the particle is
- A particle, moving in a straight line, passes through a fixed-point O. Its velocity v ms-1, t seconds after passing through O, is given by v = 6e2t – 2t.a. Find the initial velocity of the
- A particle, moving in a straight-line, passes through a fixed-price O. Its velocity v ms-1, t seconds after passing through O, is given by v = 32/(t+2)2.a. Find the acceleration of the particle when
- A particle moves in a straight line so that, ts after passing through a fixed-point O, its velocity, v ms-1, is given by v = 2t – 11 + 6/t + 1. Find the acceleration of the particle when it is at
- A particle starts from rest and moves in a straight line so that t seconds after passing through a fixed-point P, its velocity v ms-1, is given by v = 5(1 – e-t).a. Find the velocity of the
- A particle, moving in a straight-line, passes through a fixed-point O.Its velocity v ms-1, t seconds after passing through O, is given by v = 4e2t + 2t.a. Find the acceleration of the particle when t
- A particle moves in a straight line such that its displacement, x m, from a fixed-point O at time ts, is given by x = 3 + sin 2t, where t ≥ 0.a. Find the velocity of the particle when t = 0.b. Find
- A particle moves in a straight line such that its displacement, s metres, from a fixed point O on the line at time t seconds is given by s = 9[ln(3t + 2)].a. Find the value of t when the displacement
- A body moves in a straight line so that, ts after passing through a fixed-point O, its displacement from O is s m. The velocity O is sm. The velocity v ms-1 is such that v = 5 cos 4t.a. Write down
- A particle moves in a straight line so that, t seconds after passing through a fixed point O, its displacement, s metres, from O is given by s = ln(1 + 2t).a. Find the value of t when the velocity of
- A particle, moving in a straight line, passes through a fixed-point O. Its velocity v ms-1, t seconds after passing through O, is given by v = 4e2t + 6e-3t.a. Show that the velocity is never zero.b.
- A particle P is projected from the origin O so that it moves in a straight line. At time t seconds after projection, the velocity of the particle, v ms-1, is given by v = 2t2 – 14t + 12.i. Find the
- A particle travels in a straight line so that, t seconds after passing through a fixed-point O, its velocity v ms-1, is given by v = 8 cos (t/4).a. Find the value of t when the velocity of the
- A particle moves in a straight line, so that t seconds, after leaving a fixed-point O, its velocity, v ms-1, is given by v = pt2 + qt – 12, where p and q are constants.When t = 2 the acceleration
- a. Particle P moves in a straight line. Starting from rest, P moves with constant acceleration for 30 seconds after which it moves with constant velocity, k ms-1, for 90 seconds. P then moves with
- A particle, moving in a straight-line, passes through a fixed-point O. Its velocity v ms-1, t seconds after passing through O, is given by v = cos 3t + sin 3t.a. Find the value of t when the particle
- A particle moving in a straight line passes a fixed-point O with velocity 10 ms-1. Its acceleration a ms-2, t seconds after passing through O is given by a = 3 – 2t.a. Find the value of t when the
- A particle is moving in a straight line such that its velocity, v ms-1, t seconds after passing a fixed-point O is v = e2t – 6e-2t – 1.i. Find an expression for the displacement, sm, from O of
- A particle starts from rest and moves in a straight-line so that, t seconds after leaving a fixed-point O, its displacement, s metres, is given by s = 2 – 2 cos 2t.a. Find an expression for the
- A particle moving in a straight line passes a fixed-point O with velocity 18 ms-1.Its acceleration a ms-2, t seconds after passing through O is given by a = 3t – 12.a. Find the values of t when the
- A particle P is projected from the origin O so that it moves in a straight line. At time t seconds after projection, the velocity of the particle, v ms-1, is given by v = 9t2 – 63t + 90.i. Show
- A particle moves in a straight-line such that its displacement, s metres, from a fixed-point O on the line at time t seconds is given by s = 50[e-2t – e-4t].a. Find the time when the particle is
- A particle starts from rest and moves in a straight line so that, that t seconds after leaving a fixed-point O, its velocity, v ms-1, is given by v = 3 + 6 cos 2t.a. Find the range of values for the
- A particle moves in a straight line so that its displacement from a fixed-point O, given by s = 2t + 2 cos 2t, where t is the time in seconds after the motion begins.a. Find the initial position of
- A particle starts from rest at a fixed-point O and moves in a straight lie towards a point A. The velocity, v ms-1, of the particle, t seconds after leaving O, is given by v = 8 – 8e-2t.a. Find the
- A particle, moving in a straight line, passes through a fixed-point O.Its velocity vms-1, t seconds after passing through O, is given by v = k cos 4t, where k is a positive constant.a. Find the value
- A particle travels in a straight line so that, t seconds after passing through a fixed-point O, its speed, v ms-1, is given by v = 2 cos(t/2) – 1.a. Find the value of t when the particle first
- A particle moves in a straight-line such that its displacement, s metres, from a fixed point O at a time t seconds, is given byS = 2t for 0 ≤ t ≤ 4,S = 8 + 2 ln(t – 3) for t > 4.a. Find the
- A particle moves in a straight line so that t seconds after passing through a fixed-point O, its acceleration, a ms-2. Is given by a = pt + q, where p and q are constants.The particle passes through
- A particle moves in a straight line so that the displacement, s metres, from a fixed-point O, is given by s = 2t3 – 17t2 + 40t - 2, where t is the time is seconds after passing a point X on the
- A particle starts from a point O and moves in a straight line so that its displacement, s cm, from O at time t seconds is given by s = 2t sin πt/3.a. Find expression for velocity, v, and the
- The diagram shows part of the curve y = 4/2x – 1. Given that the shaded region has area 8, find the exact value of k. k
- A curve is such that dy/dx = kx2 – 2x where k is constant.Given that the curve passes through the points (1, 6) and (-2, - 15) find the equation of the curve.
- Finda. ʃ sin 4x dxb. ʃ cos 2x dxc. ʃ sin x/3 dxd. ʃ 2 cos 2x dxe. ʃ 6 sin 3x dxf. ʃ 3 cos(2x + 1) dxg. ʃ 5 sin(2 – 3x) dxh. ʃ 2 cos(2x – 7) dxi. ʃ 4 sin(1 – 5x) dx
- a. Given that y = x+2/√(3x + 4), find dy/dxb. Hence evaluates 6x + 4 dx. V(3x + 4)* 0.
- Sketch the following curves and find the area of the finite region or regions bounded by the curves and the x-axis.a. y = x(x + 1)b. y = (x + 2) (3 – x)c. y = x(x2 – 4)d. y = x(x – 2) (x + 4)e.
- Sketch the following curves and lines and find the area enclosed between their graphs.a. y = x2 + 1 and y = 5b. y = x2 – 2x + 3 and x + y = 9c. y = √x and y = 1/2xd. y = 4x – x2 and 2x + y =
- A curve is such that dy/dx = k(2x – 3)3 where k is a constant.The gradient of the normal to the curve at the point (2, 2) is – 1/8. Find the equation of the curve.
- A curve is such that dy/dx = ke2-x + 4x, where k is a constant.At the point (2, 10) the gradient of the curve is 1.a. Find the value of k.b. Find the equation of the curve.
- The point P(1, -2) lies on the curve for which dy/dx = 3 – 2/x.The point Q(2, k) also lies on the curve.a. Find the value of k.The tangents to the curve at the points P and Q intersect at the point
- A curve is such that dy/dx = 4x – 6 sin 2x.Given that the curve passes through the point (0, -2), find the equation of the curve.
- Find the exact value ofGiving your answer in the form ln(aeb), where a and b are integers. 8 5+ 4x - 3, dx,
- a. Given that y = 3(x + 1) √x – 5, show that dy/dx = 9(x – 3)/2√x-5.b. Hence finds (x – 3) dx. Jx - 5
- Evaluate.a.b.c. 3x -1)
- a. Find d/dx (x sin x).b. Hence evaluate ʃ0π/2 x cos x dx.
- a. Find the area of the region enclosed by the curve y = 12/x2, the x-axis and the lines x = 1 and x = 4.b. The line x = p divides the region in part a into two equal parts. Find the value of p.
- A curve is such that dy/dx = 2x3 + 6/x2.Given that the curve passes through the point (- 1, 10) find the equation of the curve.
- Sketch the following pairs of curves and find the area enclosed between their graphs for x ≥ 0.a. y = x2 and y = x(2 – x)b. y = x3 and y = 4x – 3x2
- A curve is such that dy/dx = 2(kx – 1)5 where k is a constant.Given that the curve passes through the points (0, 1) and (1, 8) find the equation of the curve.
- A curve is such that d2y/dx2 = 8e-2x.Given that dy/dx = 2 when x = 0 and that the curve passes through the point (1, 2/c2), find the equation of the curve.
- A curve is such that d2y/dx2 = 45 cos 3x + 2 sin x.Given that dy/dx = - 2 when x = 0 and that the curve passes through the point (π, -1), find the equation of the curve.
- a. Find the value of the constant A such that 6x-5/2x-3 = 3 + A/2x -3.b. Hence show that 6x - 5 7 dx 6+2 ln 3 s 2x-3
- a. Find d/dx(xe2x – e2x/2).b. Hence find ʃxe2x dx.
- Given thatFind the value of k. 2In 7, 3 2 1. 3x - 1
- a. Find d/dx (x2 ln x).b. Hence evaluate ʃ1e 4x ln x dx.
- a. Show that d/dx (xex – ex) xex.b. Use your result from part a to evaluate the area of the shaded region. y = x e 2
- A curve is such that dy/dx = (2 - √x)2/√x.Given that the curve passes through the point (9, 14) find the equation of the curve.
- Find the shaded area enclosed by the curve y = 3√x the line y = 10 – x and the x-axis. y = 3Jx y = 10 – x ol
- The point P(3/2, 5) lies on the curve for which dy/dx = 2e3-2x.The point Q(1, k) also lies on the curve.a. Find the value of k.The normal to the curve at the points P and Q intersect at the point
- A curve is such that dy/dx = k cos 3x – 4, where k is a constant.At the point (π, 2) the gradient of the curve is – 10.a. Find the value of k.b. Find the equation of the curve.
- a. Differentiate 4x3 ln(2x + 1) with respect to x.b. i. Given that y = 2x/√x+2, show that dy/dx = x+4/(√x+2)3.ii. Hence findiii. Hence evaluate 5x + 20 dx. (Jä + 2 )*
- a. Show that d/dx(sin x/1 – cos x) can be written in the form k/cos x – 1, and state the value of k.b. Hence find ʃ 5/cos x -1 dx.
- a. Given that 4x/2x + 3 = 2 + A/2x + 3, find the value of the constant A.b. Hence show that 4x dx 2-3 ln 2x + 3 3
- a. Given that y = x sin 3x, find dy/dx.b. Hence evaluate ʃ0π/6 x cos 3x dx.
- The tangent to the curve y = 6x – x2 at the point (2, 8) cuts the x-axis at the point P.a. Find the coordinates of P.b. Find the area of the shaded region. (2, 8) y = 6x – x 2 10 6 х
- The point (π/2, 5) lies on the curve for which dy/dx = 4 sin(2x – π/2).a. Find the equation of the curve.b. Find the equation of the normal to the curve at the point where x = π/3.
- i. Show that d/dx(e4x/4 – xe4x) = pxe4x, where p is an integer to be found.ii. Hence find the exact value of ʃ0ln2 xe4x dx, giving your answer in the form a ln 2 + b/c, where a, b and c are
- a. Given that y = (x + 8)√x – 4, show that dy/dx = kx/√x – 4, and state the value of k.b. Hence find ʃ x/√x – 4 dx.
- a. Find the quotient and remainder when 4x2 + 4x is divided by 2x + 1.b. Hence show that 4x2 + 4x dx = 2 -In 3. o 2x +1 2.
- a. Show that d/dx(x ln x) = 1 + ln x.b. Use your result from part a to evaluate the area of the shaded region. y = In x
- A curve is such that d2y/dx2 = 12x – 12.The gradient of the curve at the point (2, 9) is 8.a. Express y in terms of x.b. Show that the gradient of the curve is never less than 2.
- The curve y = √2x + 1 meets the y-axis at the point P.The tangent at the point Q(12, 5) to this curve meets the y-axis at the point R.Find the area of the shaded region POR. Q(12, 5). y=/2x + 1 R P
- The point P(π/3, 3) lies on the curve for which dy/dx = 3 cos(3x – π/2).The point Q(π/2, k) also lies on the curve.a. Find the value of k.The tangents to the curve at the points P and Q
- Find the value of1. 6x-8x dx. 2х -1
- The diagram shows part of the curve y = 3x – x3/2 and the lines y = 3x and 2y = 27 – 3x. The curve and the line y = 3x meet the x-axis at O and the curve and the line 2y = 27 – 3x meet the
- a. Given that y = 1/x2 – 7, show that dy/dx = kx/(x2 – 7)2, and state the value of k.b. Hence finds 4x dx. (x² – 7)*
- a. Show that d/dx(x cos x) = cos x – x sin x.b. Use your result from part a to evaluate the area of the shaded region. y = x sin x
- A curve is such that dy/dx = kx – 5 where k is a constant.The gradient of the normal to the curve at the point (2, -1) is – 1/3. Find the equation of the curve.
- The diagram shows the graphs of y = 2 + cos 2x and y = 1 + 2 cos 2x for 0 ≤ x ≤ π.Find the area of the shaded region. y = 2 + cos 2x TT x y = 1+ 2 cos 2x
- a. Find d/dx (2x3 ln x).b. Hence find ʃx2 ln x dx.
- The diagram shows the graphs of y = 6/x + 3 and y = 4 – x. Find the area of the shaded region. y = x + 3 y = 4 - x
- a. Differentiate x cos x with respect to x.b. Hence find ʃx sin x dx.
- a. Given that y = e2x (sin 2x + cos 2x), show that dy/dx = 4e2x cos 2x.b. Hence find ʃe2x cos 2x dx.
- a. Find d/dx(x2√2x – 7).b. Hence finds 5x2 -14x + 3 dx. V2x - 7
- Find.a. ʃ 8/x dxb. ʃ 5/x dxc. ʃ 1/2x dxd. ʃ 5/3x dxe. ʃ 1/3x+2 dxf. ʃ 1/1 – 8x dxg. ʃ 7/2x -1 dxh. ʃ 4/2 – 3x dxi. ʃ 5/2(5x – 1) dx
- a. Given that y = x + 5/√(2x – 1, show that dy/dx = x – 6/√2x – 1)3.b. Hence finds x – 6 dx. V(2x – 1)
- a. Given that y = ex2, find dy/dx.b. Use your answer to part a to find ʃxex2 dx.c. Hence evaluate ʃ20 xex2 dx.
- Evaluate.a.b.c.d.e.f.g.h.i.j.k.l. 7x° dx
- a. Given that y = (x + 1) √2x – 1, find dy/dx,b. Hence evaluates dx. 2x 1
- Find the area of the region enclosed by the curve y = 1 + cos x and the line y = 1. y = 1+ cos x -y = 1
- Find the area of each shaded region.a.b.c.d.e.f. y = x 3– 6x 2 + 9x 3
- Find y in terms of x for each of the following.a. dy/dx = 12x4b. dy/dx = 5x8c. dy/dx = 7x3d. dy/dx = 4/x3e. dy/dx = 1/2x2f. dy/dx = 3/√x
- Find each of the following.a. ʃ4x7 dxb. ʃ12x5 dxc. ʃ2x-3 dxd. ʃ 4/x2 dxe. f. 3 dx
- Find:a.b.c.d.e.f.g.h.i. |(x + 2)° dx
- Find:a. ʃe5x dxb. ʃe9x dxc. ʃe1/2x dxd. ʃe-2x dxe. ʃ4ex dxf. ʃ2e4x dxg. ʃe7x+4 dxh. ʃe5-2x dxi. ʃ1/3 e6x-1 dx
- Find:a.b.c.d.e.f.g.h.i. 7x +1+ dx
- A curve is such that dy/dx = 4x + 1/(x + 1)2 for x > 0. The curve passes through the point (1/2. 5/6).a. Find the equation of the curve.b. Find the equation of the normal to the curve at the point