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Quiz 11: One variable integration

Question

Question 1

Evaluate

\int_{0}^{1} (6 x^{2} - 4 x + 3) d x

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Your answer is correct.

\int_{0}^{1} (6 x^{2} - 4 x + 3) d x = {[2 x^{3} - 2 x^{2} + 3 x]}_{0}^{1} = 2 - 2 + 3 = 3 .

Not correct. Choice (d) is false.

Question 2

Evaluate

\int_{0}^{\frac{π}{2}} (sin x + cos x) d x

Your answer is correct.

\begin{array}{rcl} \int_{0}^{\frac{π}{2}} (sin x + cos x) d x & = & {[- cos x + sin x]}_{0}^{\frac{π}{2}} \\ = & - cos \frac{π}{2} + sin \frac{π}{2} - (- cos 0 + sin 0) \\ = & 1 + 1 = 2 . \end{array}

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Question 3

Evaluate

\int_{0}^{\frac{π}{4}} sin 2 x d x

Not correct. Choice (a) is false.

Your answer is correct.

\begin{array}{rcl} \int_{0}^{\frac{π}{4}} sin 2 x d x & = & {[- \frac{1}{2} cos 2 x]}_{0}^{\frac{π}{4}} \\ = & - \frac{1}{2} cos \frac{π}{2} + \frac{1}{2} cos 0 \\ = & \frac{1}{2} . \end{array}

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Question 4

Evaluate

\int_{0}^{\sqrt{3}} x {(1 + x^{2})}^{\frac{3}{2}} d x

Not correct. Choice (a) is false.

Your answer is correct.

\begin{array}{rcl} \int_{0}^{\sqrt{3}} x {(1 + x^{2})}^{\frac{3}{2}} d x & = & {[\frac{1}{5} {(1 + x^{2})}^{\frac{5}{2}}]}_{0}^{\sqrt{3}} \\ = & \frac{1}{5} {(4)}^{\frac{5}{2}} - \frac{1}{5} \\ = & \frac{32}{5} - \frac{1}{5} = \frac{31}{5} . \end{array}

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Question 5

Evaluate

\int_{0}^{2} 3 x^{2} e^{x^{3}} d x

Your answer is correct.

\int_{0}^{2} 3 x^{2} e^{x^{3}} d x = {[e^{x^{3}}]}_{0}^{2} = e^{8} - e^{0} = e^{8} - 1 .

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Question 6

Let

A

be the area under the curve

f (x) = 16 - x^{2}

on the interval [0,4]. Dividing the interval into 4 subintervals and defining

x_{i}

to be the midpoint of the

i

th interval (the area of the rectangle is thus

f (x_{i})

times the length of the interval), the best estimate for

A

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Your answer is correct.

The intervals are of length 1, and the area is calculated at

\frac{1}{2}

\frac{3}{2}

\frac{5}{2}

and

\frac{7}{2}

Not correct. Choice (d) is false.

Question 7

Which of the following sums is the best estimate for

A

, the area under the curve

f (x) = x^{3} + 2

on the interval [-1,2], divided into 6 subintervals and choosing

x_{i}

as the right-endpoint of the

i

th interval ?

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Your answer is correct.

The intervals are of length

\frac{1}{2}

, and the area is calculated at

- \frac{1}{2}

0

\frac{1}{2}

1

\frac{3}{2}

2

Question 8

Find the indefinite integral

\int 2 t^{2} {(1 + t^{3})}^{4} d t .

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Your answer is correct.

Let

u = 1 + t^{3}, d u = 3 t^{2} d t

\begin{array}{rcl} \int 2 t^{2} {(1 + t^{3})}^{4} d t & = & \frac{2}{3} \int u^{4} d u \\ = & \frac{2}{15} u^{5} + C \\ = & \frac{2}{15} {(1 + t^{3})}^{5} + C . \end{array}

Question 9

Find the indefinite integral

\int {sin}^{3} x cos x d x .

Not correct. Choice (a) is false.

Your answer is correct.

Let

u = sin x

d u = cos x d x .

\begin{array}{rcl} \int {sin}^{3} x cos x d x & = & \int u^{3} d u \\ = & \frac{1}{4} u^{4} + C \\ = & \frac{1}{4} {sin}^{4} x + C . \end{array}

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Question 10

Find the indefinite integral

\int \frac{2 t - e^{- t} + 4}{{(t^{2} + 4 t + e^{- t} + 1)}^{2}} d t .

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Your answer is correct.

Let

u = t^{2} + 4 t + e^{- t} + 1

d u = (2 t + 4 - e^{- t}) d t .

\begin{array}{rcl} \int \frac{2 t - e^{- t} + 4}{{(t^{2} + 4 t + e^{- t} + 1)}^{2}} d t & = & \int \frac{1}{u^{2}} d u \\ = & - u^{- 1} + C \\ = & \frac{- 1}{t^{2} + 4 t + e^{- t} + 1} + C . \end{array}

Not correct. Choice (d) is false.

ABN: 15 211 513 464. CRICOS number: 00026A. Phone: +61 2 9351 2222.

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Questions:

right first attempt
right
wrong

a)	$\frac{32}{5}$	b)	$\frac{31}{5}$
c)	$\frac{3}{2}$	d)	$\frac{9}{2}$

a)		$A = \sum_{i = 1}^{4} (16 - x_{i}^{2}) \times \frac{1}{2}$ , where $x_{i} = \frac{i - 1}{2}$
b)		$A = \sum_{i = 1}^{4} (16 - x_{i}^{2}) \times \frac{1}{2}$ , where $x_{i} = \frac{i - 1}{2}$
c)		$A = \sum_{i = 1}^{4} (16 - x_{i}^{2})$ , where $x_{i} = \frac{2 i - 1}{2}$
d)		$A = \sum_{i = 0}^{3} (16 - x_{i}^{2})$ , where $x_{i} = \frac{2 i - 1}{2}$

a)		$A = \sum_{i = 0}^{6} (x_{i}^{3} + 2)$ , where $x_{i} = \frac{i - 3}{2}$
b)		$A = \sum_{i = 0}^{6} (x_{i}^{3} + 2) \times \frac{1}{2}$ , where $x_{i} = \frac{2 i - 1}{2}$
c)		$A = \sum_{i = 1}^{6} (x_{i}^{3} + 2)$ , where $x_{i} = \frac{2 i - 3}{2}$
d)		$A = \sum_{i = 1}^{6} (x_{i}^{3} + 2) \times \frac{1}{2}$ , where $x_{i} = \frac{i - 2}{2}$

a)	$\frac{1}{5} {(1 + t^{3})}^{5} + C$	b)	$\frac{2}{5} {(1 + t^{3})}^{5} + C$
c)	$\frac{2}{3} t^{3} {(1 + t^{3})}^{5} + C$	d)	$\frac{2}{15} {(1 + t^{3})}^{5} + C$

a)	${sin}^{4} x + C$	b)	$\frac{1}{4} {sin}^{4} x + C$
c)	$3 {sin}^{2} x + C$	d)	$- \frac{1}{4} {sin}^{4} x + C$

a)	$\frac{1}{t^{2} + 4 t + e^{- t} + 1} + C$	b)	$\frac{- 1}{3 {(t^{2} + 4 t + e^{- t} + 1)}^{3}} + C$
c)	$\frac{- 1}{t^{2} + 4 t + e^{- t} + 1} + C$	d)	$\frac{1}{3 {(t^{2} + 4 t + e^{- t} + 1)}^{3}} + C$

a)	2	b)	0
c)	$\frac{π}{2}$	d)	-2

a)	1	b)	$\frac{1}{2}$
c)	$- \frac{1}{2}$	d)	-1

a)	$e^{8} - 1$	b)	$8 e^{8}$
c)	$- e^{8}$	d)	$12 e^{8} - 12$

The University of Sydney - School of Mathematics and Statistics

Quiz 11: One variable integration

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

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