Basic material

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We review basic material for this course.

Compute $9-16\cdot 0 + 8\div 2 = \answer {13}$

Find the missing quantities that complete the square: $x^2 + 20x +\answer {100} = (\answer {x+10})^2$

Find the missing quantities that complete the square: $x^2-12x + \answer {36} = (\answer {x-6})^2$

Use a change of variables to compute the following indefinite integral. $\int \frac {4x^2}{\sqrt {5 - 8x^3}} \d x$ What is the best choice of $u$ for the change of variables? $u = \answer {5 - 8x^3}$ Find $\d u$ . $\d u = \answer {-24x^2} \d x$ Rewrite the given integral using this change of variables:

$\int \frac {4x^2}{\sqrt {5 - 8x^3}} \d x = \int \answer {\frac {-1}{6} u^{-1/2}} \d u$ Find the indefinite integral: $\int \frac {4x^2}{\sqrt {5 - 8x^3}} \d x = \answer {\frac {-1}{3}(5-8x^3)^{1/2} + C}$ (Use $C$ as the arbitrary constant.)

Find the indefinite integral. $\int \frac {x}{\sqrt {x - 1}} \d x = \answer {\frac {2}{3}(x-1)^{3/2} + 2(x - 1)^{1/2} + C}$ (Use $C$ as the arbitrary constant.

Use a change of variable to evaluate the following definite integral. $\int _{-\pi /2}^{-\pi /4} -3 \cot (x) \csc ^2(x) \d x = \answer {3/2} \text {(Type an exact answer.)}$

Find a function $f$ such that $\int f(x) \d x = \cos (x^2) + C$ . $f(x) = \answer {-2 x \sin (x^2)}$

A student claims that $\int \frac {1}{1-x^2} \d x = \ln (1 - x^2) + C$ . Determine if the student is correct. If the student is incorrect, choose the best explanation that follows.

If the student’s claim had been correct, then $\dd {x} \ln (1 - x^2) = \answer {\frac {1}{1-x^2}}.$ However, $\dd {x} \ln (1 - x^2) = \answer {\frac {-2x}{1 - x^2}}$

A student is asked to compute $\int x \sqrt {1 - x^2} \d x$ on a midterm and provides the following solution: $\int x \sqrt {1 - x^2} \d x= \int x(1 -x) \d x = \int (x - x^2 )\d x = \frac {1}{2}x^2 - \frac {1}{3}x^3 + C.$ Determine if the student is correct. If the student is not correct, choose the most appropriate response below.

The mistake was in asserting that $\sqrt {1 - x^2} \ne 1 - x$ . To compute this antiderivative, the student should perform a $u$ -substitution with $u = \answer {1 - x^2}$ . Upon making this substitution, the correct antiderivative of $y = x \sqrt {1 - x^2}$ is $\answer {(-1/3)(1 - x^2)^{3/2}} + C$ .

A student is asked to compute $\int 3x^2 \cos (x) \d x$ on a midterm and provides the following solution: $\int 3x^2 \cos (x) \d x = x^3 \sin (x) + C.$ Determine if the student is correct. If the student is not correct, choose the most appropriate response below.

The mistake was in attempting to split the integral over the product. If the student were correct, then using the definition of the antiderivative, $\dd {x}(x^3\sin (x) + C) = 3x^2 \cos (x)$ . However, $\dd {x}(x^3 \sin (x) + C) = \answer {3x^2 \sin (x) + x^3 \cos (x)}$

To verify an antiderivative, differentiate and compare.

Calculate: $\int x^2 +1 \d x = \answer {x^3/3 + x+ C}$

Use $C$ for the constant.

Calculate: $\int x^3 +6 x + 1 \d x = \answer {x^4/4 + 3x^2 + x+ C}$

Use $C$ for the constant.

Calculate: $\int 3x^4 + \frac {x}{2} \d x = \answer {3x^5/5 + x^2/4 + C}$

Use $C$ for the constant.

Calculate: $\int \pi + \sin (x) \d x = \answer {\pi x - \cos (x)+ C}$

Use $C$ for the constant.

Calculate: $\int 1+ 2x + e^x \d x = \answer {x + x^2 + e^x+C}$

Use $C$ for the constant.

Calculate: $\int \frac {5}{x^5} + \frac {4}{x^4} \d x = \answer {\frac {-5}{4 x^4}-\frac {4}{3 x^3}+C}$

Use $C$ for the constant.

Calculate: $\int \frac {7}{x^6} + \frac {x^6}{7} \d x = \answer {\frac {x^7}{49}-\frac {7}{5 x^5}+C}$

Use $C$ for the constant.

Calculate: $\int \cos (x) - 2x^2 \d x = \answer {\sin (x) - 2x^3/3+C}$

Use $C$ for the constant.

Calculate: $\int e^x - 1 - x \d x = \answer {e^x-x-x^2/2+C}$

Use $C$ for the constant.

Calculate: $\int 16 \d u = \answer {16u+C}$

Use $C$ for the constant.

Calculate: $\int \d \theta = \answer {\theta +C}$

Use $C$ for the constant.

Calculate: $\int \pi \d \theta = \answer {\pi \theta +C}$

Use $C$ for the constant.

Calculate: $\int e \d t = \answer {e t + C}$

Use $C$ for the constant.

Calculate: $\int -1 \d u = \answer {-u+C}$

Use $C$ for the constant.

Calculate: $\int 7 \d z = \answer {7z+C}$

Use $C$ for the constant.

Calculate: $\int 0 \d x = \answer {0+C}$

Use $C$ for the constant.

Calculate: $\int -3 \d \theta = \answer {-3\theta +C}$

Use $C$ for the constant.

Calculate: $\int \d t = \answer {t+C}$

Use $C$ for the constant.

Calculate: $\int _0^{1/2} \frac {1}{\sqrt {1-x^2}} \d x =\answer {\frac {\pi }{6}}$

Calculate: $\int _0^1 \frac {1}{x^2 + 1} \d x = \answer {\frac {\pi }{4}}$

Calculate: $\int _0^1 \frac {1}{x + 1} \d x = \answer {\ln (2)}$

Calculate: $\int _0^{1/2} \frac {4}{\sqrt {1 - x^2}} \d x = \answer {\frac {2\pi }{3}}$

Calculate: $\int _0^{1/2} \frac {-1}{\sqrt {1 - x^2}} \d x = \answer {\frac {-\pi }{6}}$

Calculate: $\int _2^3 \sec ^2 \theta \d \theta = \answer {\sin (1) \sec (2) \sec (3)}$

Calculate: $\int _2^3 \sec (\theta )\cdot \tan (\theta ) \d \theta = \answer {\sec (3)-\sec (2)}$

Calculate: $\int _2^3 3\sec ^2\theta \d \theta = \answer {3 \sin (1) \sec (2) \sec (3)}$

Calculate: $\int _0^1 e^{-x/2} \d x = \answer {2-\frac {2}{\sqrt {e}}}$

Calculate: $\int _0^1 2e^{4x} \d x = \answer {\frac {1}{2} (e^4-1)}$

Calculate: $\int _0^3 \cos (3x) \d x = \answer {\frac {\sin (9)}{3}}$

Calculate: $\int _0^{1/2} \cos (\pi x) \d x = \answer {\frac {1}{\pi }}$

Calculate: $\int _1^2 3\sin (3x) \d x = \answer {\cos (3)-\cos (6)}$

Calculate: $\int _1^4 \pi \sin (- x) \d x = \answer {\pi (\cos (4)-\cos (1))}$

Calculate: $\int _3^5 \frac {\sin (\pi x)}{\pi } \d x = \answer {0}$

Calculate: $\int _1^2 \cos (7 x) \d x= \answer {\frac {1}{7} (\sin (14)-\sin (7))}$

Calculate: $\int _3^4 \frac {\cos (2 x)}{3} \d x=\answer {\frac {1}{6} (\sin (8)-\sin (6))}$

Calculate: $\int _0^{1/2} \pi \cos (x/\pi ) \d x = \answer {\pi ^2 \sin (\frac {1}{2 \pi })}$

Calculate: $\int _0^1 \frac {\sin (x/\pi )}{\pi } \d x = \answer {1-\cos \left (\frac {1}{\pi }\right )}$

Calculate: $\int _0^2 \pi \sin (x/\pi ) \d x =\answer {2 \pi ^2 \sin ^2\left (\frac {1}{\pi }\right )}$

Calculate: $\int _0^1 \frac {3}{(2x+1)} \d x = \answer {\frac {3 \ln (3)}{2}}$

Calculate: $\int _0^1 \frac {1}{5-3x} \d x \answer {\frac {\ln (5/2)}{3}}$

Calculate: $\int _1^2 \frac {x^3}{1 + x^4} \d x = \answer {\frac {1}{4} \ln (\frac {17}{2})}$

Calculate: $\int _0^1 \frac {2+ e^{2x}}{4x + e^{2x}} \d x = \answer {\frac {1}{2} \ln (4+e^2)}$

Calculate: $\int _0^1 \frac {1}{1 + x} \d x =\answer {\ln (2)}$

Calculate: $\int _3^6 \frac {x^2+4x^3}{x^3+3x^4} \d x =\answer {\frac {1}{3} \ln (\frac {76}{5})}$

Calculate: $\int _2^5 \frac {1}{-3-3x} \d x =\answer {\frac {-\ln (2)}{3}}$

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