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Calculus 1 : Regions

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Example Questions

Example Question #131 : Regions

What is the dot product of $a=\left \langle 5,-2,6\right \rangle$ and $b=\left \langle 1,2,8\right \rangle$ ?

Possible Answers:

Correct answer:

Explanation:

The dot product of $a=\left \langle 5,-2,6\right \rangle$ and $b=\left \langle 1,2,8\right \rangle$ is the sum of the products of its individual corresponding components, or

$(5\times 1)+(-2\times 2)+(6\times 8)=5-4+48=1+48=49$ .

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Example Question #132 : Regions

What is the dot product of $a=\left \langle 3,-3,9\right \rangle$ and $b=\left \langle 3,2,1\right \rangle$ ?

Possible Answers:

Correct answer:

Explanation:

The dot product of $a=\left \langle 3,-3,9\right \rangle$ and $b=\left \langle 3,2,1\right \rangle$ is the sum of the products of its individual corresponding components, or

$(3\times 3)+(-3\times 2)+(9\times 1)=9-6+9=3+9=12$ .

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Example Question #133 : Regions

Find the area of the region between f(x)=x+2 and g(x)=x^2-1 from x=0 to x=2 .

Possible Answers:

$5\tfrac{1}{3}$

$2\tfrac{2}{3}$

$1\tfrac{1}{3}$

Correct answer:

$5\tfrac{1}{3}$

Explanation:

To find the area between the curves, the function is

$\\A=\int_{0}^{2}(f(x)-g(x))dx\\ A=\int_{0}^{2}(x+2-(x^2-1)dx\\A=\int_{0}^{2}(x+3-x^2)dx$

then integrate from 0 to 2.

$\\A=-\frac{1}{3}\cdot 2^3+\frac{1}{2}\cdot 2^2+3\cdot 2 -\left(-\frac{1}{3}\cdot 0^3+\frac{1}{2}\cdot 0^2+3\cdot 0 \right)\\-\frac{8}{3}+2+6-0=5\frac{1}3{}$

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Example Question #51 : How To Find Area Of A Region

Find the area of the region between f(x)=2x+1 and g(x)=3x^2+2x-1 from x=0 to x=4 .

Possible Answers:

Correct answer:

Explanation:

To find the area between the curves, the function is

$\\A=\int_{0}^{4}(f(x)-g(x))dx\\ A=\int_{0}^{4}(2x+1-(3x^2+2x-1)dx\\A=\int_{0}^{4}(-3x^2+2)dx$

then integrate from 0 to 4.

A=-(4)^3+2(4)-(-0^3+2(0))=-56

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Example Question #55 : Area

Find the area of the region bound by g(x) , the -axis, and the line .

g(x)=4x^3+6x^2-8x+4

Possible Answers:

325units^2

875units^2

79.5units^2

795units^2

Correct answer:

795units^2

Explanation:

Find the area of the region bound by g(x), the x-axis, and the line .

Looks like we need to find area. Sounds like an integral problem to me.

We need a definite integral with our limits of integration at 0 and 5:

$\int_{0}^{5}g(x)dx=\int_{0}^{5}4x^3-5x^3+6x^2-8x+4dx$

Recall that to integrate polynomials we increase the exponent by 1 and divide by that number:

$\int x^n dx=\frac{x^{n+1}}{n+1}+c$

So this,

$G(x)=\int_{0}^{5}4x^3+6x^2-8x+4dx$

Becomes:

G(x)=x^4+2x^3-4x^2+4x+c_0^5

Now we need to evaluate the integral on the given interval. We do this by plugging in our limits and finding the difference:

G(5)-G(0)

Conveniently, G(0) is 0, so we really just need to find G(5)

G(5)=5^4+2(5)^3-4(5)^2+4(5)=625+250-100+20=795units^2

So, our answer is:

795units^2

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Example Question #134 : Regions

Find the area of the region between the curve of g(x) , the and axes, and the line x=4 .

g(x)=4x^3+3x^2

Possible Answers:

32 units^2

80 units^2

340 units^2

320 units^2

Correct answer:

320 units^2

Explanation:

Find the area of the region between the curve of g(x), the x and y axes, and the line

To find the area of this region, we can use a definite integral with limits of 0 and 4.

$\int_{0}^{4}4x^3+3x^2dx=x^4+x^{3}{_0^4}$

$x^4+x^{3}{_0^4}=(4^4+4^3+c)-(0^4+0^3+c)=320 units^2$

Notice that our c's cancel out and we are left with a real number for an answer.

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Example Question #135 : Regions

Find the area of the region bound by f(x) , and the -axis on the interval $\left[\frac{\pi}{2},\pi\right]$ .

{}f(x)=5sin(x)

Possible Answers:

Correct answer:

Explanation:

Find the area of the region bound by f(x), and the y-axis on the interval $\left[\frac{\pi}{2},\pi\right]$ .

{}f(x)=5sin(x)

To find the area of a region, we want to use an integral. Our limits of integration will be the endpoints of our interval.

$\int_{\frac{\pi}{2}}^{\pi}5sin(x)dx$

Recall the rule for integrating sine and use it here.

$\int_{\frac{\pi}{2}}^{\pi}5sin(x)dx=5(-cos(x)+c){_\frac{\pi}{2}^{\pi}$

Finally, evaluate the integral:

$5(-cos(\pi)+c)-5\left(-cos\left(\frac{\pi}{2}\right)+c\right)=(5+5c)-(0+5c)=5$

To get,

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Example Question #58 : Area

Find the area of the region bound by f(x) , the -axis, and the lines x=5,x=6 .

$f(x)=\frac{16}{x}$

Possible Answers:

2.917

0.182

18.23

29.17

Correct answer:

2.917

Explanation:

Find the area of the region bound by h(t), the x-axis, and the lines x=5,x=6

$f(x)=\frac{16}{x}$

To find area, set up an integral:

$\int_{5}^{6}\frac{16}{x}dx$

Recall that:

$\int\frac{1}{x}dx=ln(x)$

So we get:

$\int_{5}^{6}\frac{16}{x}dx=16ln(x)^6_5$

16ln(6)-16ln(5)=2.917

So our answer is:

2.917

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Example Question #3021 : Functions

A company produces printers that cost $\$100$ to make and sells them for $\$200$ . If the number of printers that a company sells in a year can be modeled by the equation y=-13x^2+770718x+450432 , how many printers should the company make every year in order to maximize profits?

Possible Answers:

26,753

36,326

None of the above.

31,623

29,643

Correct answer:

29,643

Explanation:

To solve this problem we must be able to find the point on the graph of the equation that is the maximum.

In order to achieve this we must take the derivative of the equation, set it equal to zero and solve for .

Taking the derivative of the equation using the power rule

$\frac{d}{dx}x^n=nx^{n-1}}$ ,

we find that the equation becomes

y'=-26x+770718 .

Setting the equation equal to 0 and solving for , we find that in order to maximize profits, the company should make 29,643 printers a year.

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Example Question #3022 : Functions

Integrate

y=2xcos(4x)

Possible Answers:

$\frac{xsin(4x)}{2}-\frac{cos(4x)}{8}+C$

$\frac{sin(4x)}{2}(x-1) +C$

$\frac{xsin(4x)}{2}+\frac{cos(4x)}{8}$

$\frac{xsin(4x)}{2}+\frac{cos(4x)}{8}+C$

$\frac{xsin(4x)}{2}-\frac{cos(4x)}{8}$

Correct answer:

$\frac{xsin(4x)}{2}+\frac{cos(4x)}{8}+C$

Explanation:

This can be integrated through integration by parts.

$u=2x\rightarrow u'=2dx$

$v'=cos(4x)\rightarrow v=\frac{sin(4x)}{4}$

$\\\int 2xcos(4x)dx\\ \\=uv-\int u'v\\ \\=\frac{xsin(4x)}{2}-\int \frac{sin(4x)dx}{2}$

$=\frac{xsin(4x)}{2}+\frac{cos(4x)}{8}+C$

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Calculus 1 : Regions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #131 : Regions

Example Question #132 : Regions

Example Question #133 : Regions

Example Question #51 : How To Find Area Of A Region

Example Question #55 : Area

Example Question #134 : Regions

Example Question #135 : Regions

Example Question #58 : Area

Example Question #3021 : Functions

Example Question #3022 : Functions

All Calculus 1 Resources

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