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Calculus 1 : Calculus

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Example Questions

Example Question #3011 : Functions

Find the area of the region under the curve of the following function:

f(x)=3x^4+9x^3+17x+1

from x=0 to x=1 .

Possible Answers:

$\frac{247}{20}$

$-\frac{247}{20}$

Correct answer:

$\frac{247}{20}$

Explanation:

The area under the curve of any function is given by the integral of the function. Note that the function evaluated from x=0 to x=1 is always positive, so the integration over the entire interval can be done using one integral. The integral of the function is given by

$\int_{0}^{1}(3x^4+9x^3+17x+1)dx$

and is equal to

$\frac{3x^5}{5}+\frac{9x^4}{4}+\frac{17x^2}{2}+x+C$ .

The integration comes from the rules

$\int u^ndu=\frac{1}{n+1}u^{n+1}+C, n\neq-1$ and $\int du=u+C$ .

Evauating the integral between the two points is done by plugging in the upper limit of integration (x=1) into the function and then subtracting by the function when the lower limit of integration (x=0) is plugged in. When evaluated from the point x=0 to x=1, the integral is equal to

$\left(\frac{3}{5}+\frac{9}{4}+\frac{17}{2}+1\right)-(0)=\frac{247}{20}$ .

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Example Question #4041 : Calculus

What is the dot product of $a=\left \langle 4,-1,3\right \rangle$ and $b=\left \langle 2,7,5\right \rangle$ ?

Possible Answers:

Correct answer:

Explanation:

The dot product of $a=\left \langle 4,-1,3\right \rangle$ and $b=\left \langle 2,7,5\right \rangle$ is the sum of the products of its individual corresponding components, or

$(4\times 2)+(-1\times 7)+(3\times 5)=8-7+15=1+15=16$ .

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

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 #51 : 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:

29.17

0.182

2.917

18.23

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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3011 : Functions

Example Question #4041 : Calculus

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

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 #51 : Area

All Calculus 1 Resources

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