Sec 4 A-Math Definite integrals and areas worksheet with answers

A free Sec 4 A-Math worksheet on definite integrals and areas: 9 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 4 A-Math worksheet: Definite integrals and areas

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

Evaluate

\int_{1}^{2} (x - 3)^{2} \, dx

Answer[2]

Find the area of the region bounded by the curve

y = x^{2} - 10x + 16

, the

x

-axis and the lines

x = 3

and

x = 5

Answer[2]

Evaluate

\int_{-1}^{2} \left(2x + 5\right) \, dx

Answer[2]

Find the area of the region enclosed by the curve

y = x^{2} + 2x - 8

and the

x

-axis.

Answer[3]

Find the area of the region enclosed by the curve

y = x^{2} - x - 1

and the line

y = 2x - 1

Answer[3]

(a)Evaluate

\int_{-1}^{2} \left(2x + 5\right) \, dx

Answer (a)[2]

(b)Evaluate

\int_{0}^{1} (3x + 1)^{2} \, dx

Answer (b)[2]

Section B

The diagram shows the curve

y = -x^{3} + 5x^{2} - 2x + 1

and the line

y = 2x + 1

. The curve and the line enclose two regions,

R_1

and

R_2

(a)Find the

x

-coordinates of the points where the line meets the curve.

Answer (a)[3]

(b)Find the total area of the two regions.

Answer (b)units²[4]

The gradient of a curve is given by

\dfrac{dy}{dx} = -2x + 2

. The curve passes through the point

(3, -3)

(a)Find the equation of the curve.

Answer (a)[3]

(b)Find the coordinates of the points where the curve meets the

x

-axis.

Answer (b)[2]

(c)Find the area of the region enclosed by the curve and the

x

-axis.

Answer (c)[3]

The diagram shows part of the curve

y = \dfrac{10}{x + 1}

. The point

P\left(4, 2\right)

lies on the curve. The tangent to the curve at

P

meets the

x

-axis at

L

. The shaded region

R

is bounded by the

y

-axis, the curve, the tangent

PL

and the

x

-axis.

(a)Find

\dfrac{dy}{dx}

Answer (a)[1]

(b)Find the equation of the tangent to the curve at

P

Answer (b)[3]

(c)Find the coordinates of

L

Answer (c)[1]

(d)Find the exact area of the shaded region

R

Answer (d)units²[4]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= \dfrac{7}{3}

[2]

\int_{1}^{2} (x - 3)^{2} \, dx = \Big[\dfrac{(x - 3)^{3}}{3}\Big]_{1}^{2}

= \left(\dfrac{((2) - 3)^{3}}{3}\right) - \left(\dfrac{((1) - 3)^{3}}{3}\right)

= \Big(-\dfrac{1}{3}\Big) - \Big(-\dfrac{8}{3}\Big)

= \dfrac{7}{3}

Question 2

Answer:

\text{Area} = \dfrac{46}{3}

units²[2]

\int_{3}^{5} \left(x^{2} - 10x + 16\right) \, dx = \Big[\dfrac{1}{3}x^{3} - 5x^{2} + 16x\Big]_{3}^{5}

= \Big(-\dfrac{10}{3}\Big) - \Big(12\Big)

= -\dfrac{46}{3}

The region is below the x-axis, so the integral is negative: the area is its size.

\text{Area} = \dfrac{46}{3}

units²

Question 3

Answer:

= 18

[2]

\int_{-1}^{2} \left(2x + 5\right) \, dx = \Big[x^{2} + 5x\Big]_{-1}^{2}

= \left((2)^{2} + 5(2)\right) - \left((-1)^{2} + 5(-1)\right)

= \Big(14\Big) - \Big(-4\Big)

= 18

Question 4

Answer:

\text{Area} = 36

units²[3]

(x + 4)(x - 2) = 0

x = -4

x = 2

\int_{-4}^{2} \left(x^{2} + 2x - 8\right) \, dx = \Big[\dfrac{1}{3}x^{3} + x^{2} - 8x\Big]_{-4}^{2}

= \Big(-\dfrac{28}{3}\Big) - \Big(\dfrac{80}{3}\Big)

= -36

The region is below the x-axis, so the integral is negative: the area is its size.

\text{Area} = 36

units²

Question 5

Answer:

= \dfrac{9}{2}

units²[3]

x^{2} - x - 1 = 2x - 1

x^{2} - 3x = 0

x(x - 3) = 0

x = 0

x = 3

Between the meeting points the line is above the curve.

\text{Area} = \int_{0}^{3} \left(-x^{2} + 3x\right) \, dx

= \Big[-\dfrac{1}{3}x^{3} + \dfrac{3}{2}x^{2}\Big]_{0}^{3}

= \Big(\dfrac{9}{2}\Big) - \Big(0\Big)

= \dfrac{9}{2}

units²

Question 6

(a)Answer:

= 18

[2]

\int_{-1}^{2} \left(2x + 5\right) \, dx = \Big[x^{2} + 5x\Big]_{-1}^{2}

= \left((2)^{2} + 5(2)\right) - \left((-1)^{2} + 5(-1)\right)

= \Big(14\Big) - \Big(-4\Big)

= 18

(b)Answer:

= 7

[2]

\int_{0}^{1} (3x + 1)^{2} \, dx = \Big[\dfrac{(3x + 1)^{3}}{9}\Big]_{0}^{1}

= \left(\dfrac{(3(1) + 1)^{3}}{9}\right) - \left(\dfrac{(3(0) + 1)^{3}}{9}\right)

= \Big(\dfrac{64}{9}\Big) - \Big(\dfrac{1}{9}\Big)

= 7

Question 7

(a)Answer:

x = 0, 1, 4

[3]

-x^{3} + 5x^{2} - 2x + 1 = 2x + 1

-x^{3} + 5x^{2} - 4x = 0

-x(x - 1)(x - 4) = 0

x = 0, 1

4

(b)Answer:

\dfrac{71}{6}

units²[4]

Curve − line

= -x^{3} + 5x^{2} - 4x

\displaystyle\int_{0}^{1} (\text{curve} - \text{line}) \, dx = -\dfrac{7}{12}

\displaystyle\int_{1}^{4} (\text{curve} - \text{line}) \, dx = \dfrac{45}{4}

Total area

= \dfrac{7}{12} + \dfrac{45}{4}

= \dfrac{71}{6}

units²

Question 8

(a)Answer:

y = -x^{2} + 2x

[3]

y = \int (-2x + 2) \, dx

= -x^{2} + 2x + c

-3 = -3 + c

c = 0

y = -x^{2} + 2x

(b)Answer:

x = 0

x = 2

[2]

-x^{2} + 2x = 0

-x(x - 2) = 0

x = 0

x = 2

(c)Answer:

\dfrac{4}{3}

units²[3]

\int_{0}^{2} (-x^{2} + 2x) \, dx = \left[-\dfrac{1}{3}x^{3} + x^{2}\right]_{0}^{2}

= \dfrac{4}{3}

Area

= \dfrac{4}{3}

units²

Question 9

(a)Answer:

\dfrac{dy}{dx} = -\dfrac{10}{(x + 1)^{2}}

[1]

y = 10(x + 1)^{-1}

\dfrac{dy}{dx} = -10(x + 1)^{-2} \times 1

= -\dfrac{10}{(x + 1)^{2}}

(b)Answer:

y = -\dfrac{2}{5}x + \dfrac{18}{5}

[3]

x = 4

\dfrac{dy}{dx} = -\dfrac{10}{5^{2}}

\dfrac{dy}{dx} = -\dfrac{2}{5}

y - 2 = -\dfrac{2}{5}(x - 4)

y = -\dfrac{2}{5}x + \dfrac{18}{5}

(c)Answer:

L\left(9, 0\right)

[1]

L

y = 0

0 = -\dfrac{2}{5}x + \dfrac{18}{5}

x = 9

(d)Answer:

10\ln 5 + 5

units²[4]

Area under the curve from

x = 0

x = 4

\int_{0}^{4} \dfrac{10}{x + 1} \, dx = \left[10\ln(x + 1)\right]_{0}^{4}

= 10\ln 5 - 10\ln 1

= 10\ln 5

Area of the triangle under

PL

= \dfrac{1}{2} \times \left(9 - 4\right) \times 2

= 5

Area of

R = 10\ln 5 + 5

Practice questions by skill

Sec 4 A-Math Definite integrals and areas worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Definite integrals and areas notes. Or practise it online with step-by-step help when you get one wrong.

Section B

The diagram shows the curve

y = -x^{3} + 5x^{2} - 2x + 1

and the line

y = 2x + 1

. The curve and the line enclose two regions,

R_1

and

R_2

(a)Find the

x

-coordinates of the points where the line meets the curve.

Answer (a)[3]

(b)Find the total area of the two regions.

Answer (b)units²[4]

Answers and worked solutions

Question 1

Answer:

= \dfrac{7}{3}

[2]

\int_{1}^{2} (x - 3)^{2} \, dx = \Big[\dfrac{(x - 3)^{3}}{3}\Big]_{1}^{2}

= \left(\dfrac{((2) - 3)^{3}}{3}\right) - \left(\dfrac{((1) - 3)^{3}}{3}\right)

= \Big(-\dfrac{1}{3}\Big) - \Big(-\dfrac{8}{3}\Big)

= \dfrac{7}{3}

Question 2

Answer:

\text{Area} = \dfrac{46}{3}

units²[2]

\int_{3}^{5} \left(x^{2} - 10x + 16\right) \, dx = \Big[\dfrac{1}{3}x^{3} - 5x^{2} + 16x\Big]_{3}^{5}

= \Big(-\dfrac{10}{3}\Big) - \Big(12\Big)

= -\dfrac{46}{3}

The region is below the x-axis, so the integral is negative: the area is its size.

\text{Area} = \dfrac{46}{3}

units²

Question 3

Answer:

= 18

[2]

\int_{-1}^{2} \left(2x + 5\right) \, dx = \Big[x^{2} + 5x\Big]_{-1}^{2}

= \left((2)^{2} + 5(2)\right) - \left((-1)^{2} + 5(-1)\right)

= \Big(14\Big) - \Big(-4\Big)

= 18

Question 4

Answer:

\text{Area} = 36

units²[3]

(x + 4)(x - 2) = 0

x = -4

x = 2

\int_{-4}^{2} \left(x^{2} + 2x - 8\right) \, dx = \Big[\dfrac{1}{3}x^{3} + x^{2} - 8x\Big]_{-4}^{2}

= \Big(-\dfrac{28}{3}\Big) - \Big(\dfrac{80}{3}\Big)

= -36

The region is below the x-axis, so the integral is negative: the area is its size.

\text{Area} = 36

units²

Question 5

Answer:

= \dfrac{9}{2}

units²[3]

x^{2} - x - 1 = 2x - 1

x^{2} - 3x = 0

x(x - 3) = 0

x = 0

x = 3

Between the meeting points the line is above the curve.

\text{Area} = \int_{0}^{3} \left(-x^{2} + 3x\right) \, dx

= \Big[-\dfrac{1}{3}x^{3} + \dfrac{3}{2}x^{2}\Big]_{0}^{3}

= \Big(\dfrac{9}{2}\Big) - \Big(0\Big)

= \dfrac{9}{2}

units²

Question 6

(a)Answer:

= 18

[2]

\int_{-1}^{2} \left(2x + 5\right) \, dx = \Big[x^{2} + 5x\Big]_{-1}^{2}

= \left((2)^{2} + 5(2)\right) - \left((-1)^{2} + 5(-1)\right)

= \Big(14\Big) - \Big(-4\Big)

= 18

(b)Answer:

= 7

[2]

\int_{0}^{1} (3x + 1)^{2} \, dx = \Big[\dfrac{(3x + 1)^{3}}{9}\Big]_{0}^{1}

= \left(\dfrac{(3(1) + 1)^{3}}{9}\right) - \left(\dfrac{(3(0) + 1)^{3}}{9}\right)

= \Big(\dfrac{64}{9}\Big) - \Big(\dfrac{1}{9}\Big)

= 7

Question 7

(a)Answer:

x = 0, 1, 4

[3]

-x^{3} + 5x^{2} - 2x + 1 = 2x + 1

-x^{3} + 5x^{2} - 4x = 0

-x(x - 1)(x - 4) = 0

x = 0, 1

4

(b)Answer:

\dfrac{71}{6}

units²[4]

Curve − line

= -x^{3} + 5x^{2} - 4x

\displaystyle\int_{0}^{1} (\text{curve} - \text{line}) \, dx = -\dfrac{7}{12}

\displaystyle\int_{1}^{4} (\text{curve} - \text{line}) \, dx = \dfrac{45}{4}

Total area

= \dfrac{7}{12} + \dfrac{45}{4}

= \dfrac{71}{6}

units²

Question 8

(a)Answer:

y = -x^{2} + 2x

[3]

y = \int (-2x + 2) \, dx

= -x^{2} + 2x + c

-3 = -3 + c

c = 0

y = -x^{2} + 2x

(b)Answer:

x = 0

x = 2

[2]

-x^{2} + 2x = 0

-x(x - 2) = 0

x = 0

x = 2

(c)Answer:

\dfrac{4}{3}

units²[3]

\int_{0}^{2} (-x^{2} + 2x) \, dx = \left[-\dfrac{1}{3}x^{3} + x^{2}\right]_{0}^{2}

= \dfrac{4}{3}

Area

= \dfrac{4}{3}

units²

Question 9

(a)Answer:

\dfrac{dy}{dx} = -\dfrac{10}{(x + 1)^{2}}

[1]

y = 10(x + 1)^{-1}

\dfrac{dy}{dx} = -10(x + 1)^{-2} \times 1

= -\dfrac{10}{(x + 1)^{2}}

(b)Answer:

y = -\dfrac{2}{5}x + \dfrac{18}{5}

[3]

x = 4

\dfrac{dy}{dx} = -\dfrac{10}{5^{2}}

\dfrac{dy}{dx} = -\dfrac{2}{5}

y - 2 = -\dfrac{2}{5}(x - 4)

y = -\dfrac{2}{5}x + \dfrac{18}{5}

(c)Answer:

L\left(9, 0\right)

[1]

L

y = 0

0 = -\dfrac{2}{5}x + \dfrac{18}{5}

x = 9

(d)Answer:

10\ln 5 + 5

units²[4]

Area under the curve from

x = 0

x = 4

\int_{0}^{4} \dfrac{10}{x + 1} \, dx = \left[10\ln(x + 1)\right]_{0}^{4}

= 10\ln 5 - 10\ln 1

= 10\ln 5

Area of the triangle under

PL

= \dfrac{1}{2} \times \left(9 - 4\right) \times 2

= 5

Area of

R = 10\ln 5 + 5