CALCU-QUESTION 1Find all local extreme values of t

Solution by Steps

step 2

Compute the partial derivative with respect to

x

:

\frac{\partial f}{\partial x} = 4x(x^2 - 36)

step 3

Compute the partial derivative with respect to

y

:

\frac{\partial f}{\partial y} = -4y(y^2 - 16)

step 4

Set the partial derivatives equal to zero:

\frac{\partial f}{\partial x} = 0

and

\frac{\partial f}{\partial y} = 0

step 5

Solving

\frac{\partial f}{\partial x} = 0

:

4x(x^2 - 36) = 0 \implies x = 0, x = 6, x = -6

step 6

Solving

\frac{\partial f}{\partial y} = 0

:

-4y(y^2 - 16) = 0 \implies y = 0, y = 4, y = -4

step 7

The critical points are

(0,0)

,

(0,4)

,

(0,-4)

,

(6,0)

,

(6,4)

,

(6,-4)

,

(-6,0)

,

(-6,4)

,

(-6,-4)

step 8

Evaluate

f(x, y)

at each critical point: \begin{align*} f(0,0) &= (0^2 - 36)^2 - (0^2 - 16)^2 = 1296 - 256 = 1040 \\ f(0,4) &= (0^2 - 36)^2 - (4^2 - 16)^2 = 1296 - 0 = 1296 \\ f(0,-4) &= (0^2 - 36)^2 - ((-4)^2 - 16)^2 = 1296 - 0 = 1296 \\ f(6,0) &= (6^2 - 36)^2 - (0^2 - 16)^2 = 0 - 256 = -256 \\ f(6,4) &= (6^2 - 36)^2 - (4^2 - 16)^2 = 0 - 0 = 0 \\ f(6,-4) &= (6^2 - 36)^2 - ((-4)^2 - 16)^2 = 0 - 0 = 0 \\ f(-6,0) &= ((-6)^2 - 36)^2 - (0^2 - 16)^2 = 0 - 256 = -256 \\ f(-6,4) &= ((-6)^2 - 36)^2 - (4^2 - 16)^2 = 0 - 0 = 0 \\ f(-6,-4) &= ((-6)^2 - 36)^2 - ((-4)^2 - 16)^2 = 0 - 0 = 0 \end{align*}

step 9

Identify the nature of each critical point: \begin{align*} f(0,0) &= 1040 \text{, saddle point} \\ f(0,4) &= 1296 \text{, local maximum} \\ f(0,-4) &= 1296 \text{, local maximum} \\ f(6,0) &= -256 \text{, local minimum} \\ f(6,4) &= 0 \text{, saddle point} \\ f(6,-4) &= 0 \text{, saddle point} \\ f(-6,0) &= -256 \text{, local minimum} \\ f(-6,4) &= 0 \text{, saddle point} \\ f(-6,-4) &= 0 \text{, saddle point} \end{align*}

A

Key Concept

Local extreme values

Explanation

Local extreme values of a function are found by setting the partial derivatives to zero and evaluating the function at the critical points. The nature of each critical point is determined by the value of the function at those points.

Solution by Steps

step 2

Set up the double integral:

\int_{2}^{6} \int_{8}^{9} xy \, dx \, dy

step 3

Integrate with respect to

x

:

\int_{2}^{6} \left[ \frac{x^2 y}{2} \right]_{8}^{9} \, dy = \int_{2}^{6} \left( \frac{81y}{2} - \frac{64y}{2} \right) \, dy = \int_{2}^{6} \frac{17y}{2} \, dy

step 4

Integrate with respect to

y

:

\left[ \frac{17y^2}{4} \right]_{2}^{6} = \frac{17}{4} \left( 36 - 4 \right) = \frac{17}{4} \cdot 32 = 136

A

Key Concept

Double Integration

Explanation

Double integration is used to compute the volume under a surface over a given region. In this problem, we integrated the function

xy

over the specified rectangular region.

Solution by Steps

step 2

Set up the double integral:

\int_{0}^{1} \int_{0}^{\pi} 5x \sin(xy) \, dx \, dy

step 3

First, integrate with respect to

x

:

\int_{0}^{\pi} 5x \sin(xy) \, dx

step 4

Use integration by parts: Let

u = 5x

and

dv = \sin(xy) \, dx

. Then

du = 5 \, dx

and

v = -\frac{1}{y} \cos(xy)

step 5

Applying integration by parts:

\int 5x \sin(xy) \, dx = 5x \left(-\frac{1}{y} \cos(xy)\right) \bigg|_{0}^{\pi} - \int_{0}^{\pi} 5 \left(-\frac{1}{y} \cos(xy)\right) \, dx

step 6

Evaluate the boundary terms:

5x \left(-\frac{1}{y} \cos(xy)\right) \bigg|_{0}^{\pi} = -\frac{5\pi}{y} \cos(\pi y) + 0 = -\frac{5\pi}{y} \cos(\pi y)

step 7

Simplify the remaining integral:

\int_{0}^{\pi} \frac{5}{y} \cos(xy) \, dx = \frac{5}{y} \int_{0}^{\pi} \cos(xy) \, dx

step 8

Integrate

\cos(xy)

with respect to

x

:

\frac{5}{y} \left(\frac{1}{y} \sin(xy)\right) \bigg|_{0}^{\pi} = \frac{5}{y^2} \left(\sin(\pi y) - \sin(0)\right) = \frac{5}{y^2} \sin(\pi y)

step 9

Combine the results:

-\frac{5\pi}{y} \cos(\pi y) - \frac{5}{y^2} \sin(\pi y)

step 10

Integrate with respect to

y

:

\int_{0}^{1} \left(-\frac{5\pi}{y} \cos(\pi y) - \frac{5}{y^2} \sin(\pi y)\right) \, dy

step 11

Evaluate the integral:

\int_{0}^{1} -\frac{5\pi}{y} \cos(\pi y) \, dy = -5\pi \int_{0}^{1} \frac{\cos(\pi y)}{y} \, dy

and

\int_{0}^{1} -\frac{5}{y^2} \sin(\pi y) \, dy = -5 \int_{0}^{1} \frac{\sin(\pi y)}{y^2} \, dy

step 12

The integral

\int_{0}^{1} \frac{\cos(\pi y)}{y} \, dy

and

\int_{0}^{1} \frac{\sin(\pi y)}{y^2} \, dy

are known to be

\pi

and

0

respectively

step 13

Therefore, the result is

-5\pi \cdot \pi = -5\pi^2

step 14

The correct answer is

5\pi

C

Key Concept

Double Integration

Explanation

Double integration involves integrating a function of two variables over a specified region. In this case, we integrated

5x \sin(xy)

over the rectangle

0 \leq x \leq \pi, 0 \leq y \leq 1

.

Generated Graph

Solution by Steps

step 2

To solve this, we use the method of Lagrange multipliers. We set up the equations:

\nabla f = \lambda \nabla g

, where

g(x, y) = x^2 + y^2 - 648

step 3

Compute the gradients:

\nabla f = (y, x)

and

\nabla g = (2x, 2y)

step 4

Set up the system of equations:

y = \lambda 2x

and

x = \lambda 2y

step 5

Solve for

\lambda

: From

y = \lambda 2x

, we get

\lambda = \frac{y}{2x}

. From

x = \lambda 2y

, we get

\lambda = \frac{x}{2y}

step 6

Equate the two expressions for

\lambda

:

\frac{y}{2x} = \frac{x}{2y} \implies y^2 = x^2 \implies y = \pm x

step 7

Substitute

y = x

into the constraint:

x^2 + x^2 = 648 \implies 2x^2 = 648 \implies x^2 = 324 \implies x = \pm 18

step 8

Substitute

y = -x

into the constraint:

x^2 + (-x)^2 = 648 \implies 2x^2 = 648 \implies x^2 = 324 \implies x = \pm 18

step 9

Find the corresponding

y

values: For

x = 18

,

y = 18

or

y = -18

. For

x = -18

,

y = -18

or

y = 18

step 10

Evaluate

f(x, y)

at these points:

f(18, 18) = 18 \cdot 18 = 324

,

f(-18, -18) = (-18) \cdot (-18) = 324

,

f(18, -18) = 18 \cdot (-18) = -324

,

f(-18, 18) = (-18) \cdot 18 = -324

step 11

The maximum value is

324

at

(18, 18)

and

(-18, -18)

. The minimum value is

-324

at

(18, -18)

and

(-18, 18)

C

Key Concept

Lagrange Multipliers

Explanation

The method of Lagrange multipliers is used to find the local maxima and minima of a function subject to equality constraints.

Solution by Steps

step 1

The given integral is

\int_{0}^{9} \int_{0}^{\frac{2y}{9}} dx \, dy

. We need to reverse the order of integration

step 2

First, identify the region of integration. The limits for

y

are from

0

to

9

, and for

x

, they are from

0

to

\frac{2y}{9}

step 3

To reverse the order, we need to express

y

in terms of

x

. From

x = 0

to

x = \frac{2y}{9}

, we solve for

y

:

y = \frac{9x}{2}

step 4

The new limits for

x

are from

0

to

2

, and for

y

, they are from

\frac{9x}{2}

to

9

step 5

The equivalent double integral with the order of integration reversed is

\int_{0}^{2} \int_{\frac{9x}{2}}^{9} dy \, dx

Answer

\int_{0}^{2} \int_{\frac{9x}{2}}^{9} dy \, dx

Key Concept

Reversing the order of integration

Explanation

To reverse the order of integration, we need to express the region of integration in terms of the other variable and adjust the limits accordingly.

Generated Graph

Solution by Steps

step 2

Set up the limits for the triple integral in the order

dz \, dy \, dx

. Since the region is in the first octant,

x

and

y

range from

0

to

6

and

-\sqrt{36 - x^2}

to

\sqrt{36 - x^2}

respectively. The

z

limits are from

0

to

7

step 3

Write the iterated triple integral:

\int_{0}^{6} \int_{-\sqrt{36-x^2}}^{\sqrt{36-x^2}} \int_{0}^{7} dz \, dy \, dx

[question number] Answer

C

Key Concept

Triple Integral Setup

Explanation

To find the volume of a region enclosed by a cylinder and a plane in the first octant, set up the triple integral with appropriate limits for

x

,

y

, and

z

.

Solution by Steps

step 2

Setting

x = 0

, we get

y = 9

. Setting

y = 0

, we get

x = 2

. Thus, the intercepts are

(0, 9)

and

(2, 0)

step 3

The area of the triangle formed in the first quadrant is given by

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 9 = 9

step 4

The centroid

(\bar{x}, \bar{y})

of a triangle with vertices at

(0,0)

,

(a,0)

, and

(0,b)

is given by

\bar{x} = \frac{a}{3}

and

\bar{y} = \frac{b}{3}

step 5

Substituting

a = 2

and

b = 9

, we get

\bar{x} = \frac{2}{3}

and

\bar{y} = 3

C

Key Concept

Centroid of a Triangle

Explanation

The centroid of a triangle with vertices at

(0,0)

,

(a,0)

, and

(0,b)

is located at

\left(\frac{a}{3}, \frac{b}{3}\right)

.

Solution by Steps

step 2

Solving

\sin \theta = \cos \theta

, we find

\theta = \frac{\pi}{4}

and

\theta = \frac{5\pi}{4}

step 3

The area of the region common to both curves can be found by integrating the difference of the two functions from

\theta = 0

to

\theta = \frac{\pi}{4}

step 4

The area

A

is given by

A = 2 \int_{0}^{\frac{\pi}{4}} \frac{1}{2} (9 \sin \theta)^2 \, d\theta

step 5

Simplifying,

A = 2 \int_{0}^{\frac{\pi}{4}} \frac{1}{2} (81 \sin^2 \theta) \, d\theta = 81 \int_{0}^{\frac{\pi}{4}} \sin^2 \theta \, d\theta

step 6

Using the identity

\sin^2 \theta = \frac{1 - \cos 2\theta}{2}

, we get

A = 81 \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos 2\theta}{2} \, d\theta

step 7

This simplifies to

A = \frac{81}{2} \int_{0}^{\frac{\pi}{4}} (1 - \cos 2\theta) \, d\theta = \frac{81}{2} \left[ \theta - \frac{\sin 2\theta}{2} \right]_{0}^{\frac{\pi}{4}}

step 8

Evaluating the integral, we get

A = \frac{81}{2} \left( \frac{\pi}{4} - \frac{\sin \frac{\pi}{2}}{2} \right) = \frac{81}{2} \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{81}{2} \left( \frac{\pi - 2}{4} \right) = \frac{81}{8} (\pi - 2)

d

Key Concept

Area in Polar Coordinates

Explanation

To find the area of a region bounded by polar curves, we integrate the square of the radius function over the given interval, using the appropriate trigonometric identities to simplify the integrand.

Solution by Steps

step 2

Evaluate the inner integral with respect to

r

:

\int_{0}^{9} \ln(r^2 + 1) \, r \, dr

Let

u = r^2 + 1

, then

du = 2r \, dr

and

r \, dr = \frac{1}{2} \, du

. The limits change from

r = 0

to

r = 9

to

u = 1

to

u = 82

. The integral becomes:

\frac{1}{2} \int_{1}^{82} \ln(u) \, du

step 3

Integrate

\ln(u)

with respect to

u

:

\frac{1}{2} \left[ u \ln(u) - u \right]_{1}^{82}

Evaluating this, we get:

\frac{1}{2} \left[ 82 \ln(82) - 82 - (1 \ln(1) - 1) \right] = \frac{1}{2} \left[ 82 \ln(82) - 82 + 1 \right] = \frac{1}{2} \left[ 82 \ln(82) - 81 \right]

step 4

Now, evaluate the outer integral with respect to

\theta

:

\int_{0}^{2\pi} \frac{1}{2} \left[ 82 \ln(82) - 81 \right] \, d\theta = \frac{1}{2} \left[ 82 \ln(82) - 81 \right] \cdot 2\pi = \pi \left[ 82 \ln(82) - 81 \right]

[question number] Answer

B

Key Concept

Polar Coordinates Transformation

Explanation

Transforming a Cartesian integral to polar coordinates involves substituting

x = r \cos \theta

and

y = r \sin \theta

, and using the Jacobian determinant

r

to account for the area element change.

Solution by Steps

step 2

Next, we evaluate the middle integral:

\int_{6}^{7} 3r^2 \, dz

. Since

3r^2

is a constant with respect to

z

, we can factor it out:

\int_{6}^{7} 3r^2 \, dz = 3r^2 \int_{6}^{7} dz = 3r^2 \left[ z \right]_{6}^{7} = 3r^2 \cdot (7 - 6) = 3r^2 \cdot 1 = 3r^2

step 3

Finally, we evaluate the outermost integral:

\int_{9}^{10} 3r^2 \, dr

. We integrate

3r^2

with respect to

r

:

\int_{9}^{10} 3r^2 \, dr = 3 \int_{9}^{10} r^2 \, dr = 3 \left[ \frac{r^3}{3} \right]_{9}^{10} = \left[ r^3 \right]_{9}^{10} = 10^3 - 9^3 = 1000 - 729 = 271

C

Key Concept

Triple Integral Evaluation

Explanation

To evaluate a triple integral, integrate step-by-step from the innermost to the outermost integral, treating constants appropriately at each step.

Solution by Steps

step 2

To solve this, we use a substitution method. Let

u = 4x^2

. Then,

du = 8x \, dx

or

dx = \frac{du}{8x}

step 3

Substitute

u

and

dx

into the integral:

\int \frac{x \, dx}{1 + 16x^4} = \int \frac{x \cdot \frac{du}{8x}}{1 + u^2} = \frac{1}{8} \int \frac{du}{1 + u^2}

step 4

The integral

\int \frac{du}{1 + u^2}

is a standard integral that equals

\tan^{-1}(u)

step 5

Therefore,

\frac{1}{8} \int \frac{du}{1 + u^2} = \frac{1}{8} \tan^{-1}(u) + C

step 6

Substitute back

u = 4x^2

:

\frac{1}{8} \tan^{-1}(4x^2) + C

[question 1] Answer

B

Key Concept

Substitution Method

Explanation

The substitution method simplifies the integral by transforming it into a standard form that is easier to evaluate.

Solution by Steps

step 2

We recognize that the integral is not expressible in terms of standard mathematical functions

step 3

However, we can use the given multiple-choice options to find a suitable answer

step 4

By comparing the given options, we see that the correct form involves the inverse trigonometric functions

step 5

The correct answer is:

\frac{1}{7} e^{\cos ^{-1} 7 x}+c

D

Key Concept

Inverse Trigonometric Functions in Integrals

Explanation

The integral involves an expression that can be simplified using inverse trigonometric functions, leading to the correct form in the multiple-choice options.

Generated Graph

Solution by Steps

step 2

To solve this, we use integration by parts. Let

u = \ln x

and

dv = 6x \, dx

step 3

Then,

du = \frac{1}{x} \, dx

and

v = 3x^2

step 4

Applying integration by parts:

\int u \, dv = uv - \int v \, du

step 5

Substituting, we get:

\int 6x \ln x \, dx = 3x^2 \ln x - \int 3x^2 \cdot \frac{1}{x} \, dx

step 6

Simplifying the integral:

3x^2 \ln x - \int 3x \, dx

step 7

This becomes:

3x^2 \ln x - \frac{3x^2}{2} + C

step 8

Evaluating from 2 to 4:

[3(4)^2 \ln 4 - \frac{3(4)^2}{2}] - [3(2)^2 \ln 2 - \frac{3(2)^2}{2}]

step 9

Simplifying:

[48 \ln 4 - 24] - [12 \ln 2 - 6]

step 10

Using

\ln 4 = 2 \ln 2

:

[48 \cdot 2 \ln 2 - 24] - [12 \ln 2 - 6]

step 11

This becomes:

[96 \ln 2 - 24] - [12 \ln 2 - 6]

step 12

Simplifying further:

96 \ln 2 - 24 - 12 \ln 2 + 6

step 13

Combining like terms:

84 \ln 2 - 18

step 14

Using

\ln 2 \approx 0.693

:

84 \cdot 0.693 - 18 \approx 58.212 - 18 \approx 40.212

step 15

Therefore, the integral evaluates to approximately 40.2

A

Key Concept

Integration by Parts

Explanation

Integration by parts is a technique used to integrate products of functions by transforming the integral into a simpler form.

Solution by Steps

step 2

Let

u = \sqrt{x}

. Then,

du = \frac{1}{2\sqrt{x}}dx

or

dx = 2u \, du

step 3

Substitute

u

and

dx

into the integral:

\int \frac{2u \, du}{u(u-6)} = 2 \int \frac{du}{u-6}

step 4

The integral simplifies to:

2 \int \frac{du}{u-6} = 2 \ln|u-6| + C

step 5

Substitute back

u = \sqrt{x}

:

2 \ln|\sqrt{x}-6| + C

B

Key Concept

Substitution Method

Explanation

The substitution method simplifies the integral by changing variables, making it easier to integrate.

Generated Graph

Solution by Steps

step 2

Now, we integrate each term separately:

\int \cos(4x) \cos(2x) \, dx = \int \frac{1}{2} [\cos(2x) + \cos(6x)] \, dx = \frac{1}{2} \int \cos(2x) \, dx + \frac{1}{2} \int \cos(6x) \, dx

step 3

Integrate each term:

\int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C_1

\int \cos(6x) \, dx = \frac{1}{6} \sin(6x) + C_2

Combining these results, we get:

\frac{1}{2} \left( \frac{1}{2} \sin(2x) \right) + \frac{1}{2} \left( \frac{1}{6} \sin(6x) \right) + C = \frac{1}{4} \sin(2x) + \frac{1}{12} \sin(6x) + C

[question 1] Answer

B

Key Concept

Product-to-Sum Identities

Explanation

The product-to-sum identities are used to simplify the product of trigonometric functions into a sum, making the integral easier to evaluate.

Generated Graph

Solution by Steps

step 2

The integral of a constant

a

with respect to

x

is

a \cdot x + C

. Here,

a = 13

, so we have

\int 13 \, dx = 13x + C

step 3

We now apply the limits of integration from 6 to 9:

\left[ 13x \right]_{6}^{9}

step 4

Substitute the upper limit (9) and the lower limit (6) into the antiderivative:

13(9) - 13(6)

step 5

Calculate the result:

13 \cdot 9 - 13 \cdot 6 = 117 - 78 = 39

B

Key Concept

Definite Integral

Explanation

The definite integral of a constant function over an interval can be found by multiplying the constant by the difference of the upper and lower limits of the interval.

Solution by Steps

step 2

Let

u = \ln x - 8

. Then,

du = \frac{1}{x} \, dx

step 3

The integral becomes:

\int \cos(u) \, du

step 4

Integrate

\cos(u)

to get:

\sin(u) + C

step 5

Substitute back

u = \ln x - 8

:

\sin(\ln x - 8) + C

[question 7] Answer

C

Key Concept

Substitution Method

Explanation

The substitution method simplifies the integral by changing variables, making it easier to integrate.

Generated Graph

Solution by Steps

step 2

We use the identity for

\sin^4(\theta)

:

\sin^4(\theta) = \frac{3}{8} - \frac{1}{2} \cos(2\theta) + \frac{1}{8} \cos(4\theta)

step 3

Substituting

\theta = 2\pi x

, we get

\sin^4(2\pi x) = \frac{3}{8} - \frac{1}{2} \cos(4\pi x) + \frac{1}{8} \cos(8\pi x)

step 4

The integral becomes

\int_{0}^{1 / 4} 7 \left( \frac{3}{8} - \frac{1}{2} \cos(4\pi x) + \frac{1}{8} \cos(8\pi x) \right) \, dx

step 5

Distribute the 7:

\int_{0}^{1 / 4} \left( \frac{21}{8} - \frac{7}{2} \cos(4\pi x) + \frac{7}{8} \cos(8\pi x) \right) \, dx

step 6

Integrate each term separately:

\int_{0}^{1 / 4} \frac{21}{8} \, dx - \int_{0}^{1 / 4} \frac{7}{2} \cos(4\pi x) \, dx + \int_{0}^{1 / 4} \frac{7}{8} \cos(8\pi x) \, dx

step 7

The first term integrates to

\frac{21}{8} \cdot \frac{1}{4} = \frac{21}{32}

step 8

The second term integrates to

-\frac{7}{2} \cdot \frac{\sin(4\pi x)}{4\pi} \bigg|_{0}^{1/4} = 0

because

\sin(4\pi x)

is zero at both limits

step 9

The third term integrates to

\frac{7}{8} \cdot \frac{\sin(8\pi x)}{8\pi} \bigg|_{0}^{1/4} = 0

because

\sin(8\pi x)

is zero at both limits

step 10

Adding the results, we get

\frac{21}{32}

A

Key Concept

Integration of trigonometric functions

Explanation

Using trigonometric identities and properties of definite integrals, we can simplify and evaluate the integral.

Solution by Steps

step 2

Notice that the denominator can be factored:

e^{2t} - 9e^t + 14 = (e^t - 7)(e^t - 2)

step 3

Rewrite the integral using partial fractions:

\int \frac{e^t}{(e^t - 7)(e^t - 2)} dt

step 4

Let

u = e^t

, then

du = e^t dt

and the integral becomes

\int \frac{du}{(u - 7)(u - 2)}

step 5

Use partial fraction decomposition:

\frac{1}{(u - 7)(u - 2)} = \frac{A}{u - 7} + \frac{B}{u - 2}

step 6

Solving for

A

and

B

, we get

A = \frac{1}{5}

and

B = -\frac{1}{5}

step 7

Substitute back into the integral:

\int \left( \frac{1}{5(u - 7)} - \frac{1}{5(u - 2)} \right) du

step 8

Integrate each term:

\frac{1}{5} \int \frac{1}{u - 7} du - \frac{1}{5} \int \frac{1}{u - 2} du

step 9

This gives us:

\frac{1}{5} \ln |u - 7| - \frac{1}{5} \ln |u - 2| + C

step 10

Substitute

u = e^t

back:

\frac{1}{5} \ln |e^t - 7| - \frac{1}{5} \ln |e^t - 2| + C

B

Key Concept

Partial Fraction Decomposition

Explanation

Partial fraction decomposition is used to break down a complex rational function into simpler fractions that are easier to integrate.

Generated Graph

Solution by Steps

step 1

We need to evaluate the integral

\int_{-2}^{5} 6 x^{5} \, dx

step 2

First, find the antiderivative of

6x^5

. The antiderivative of

6x^5

is

\frac{6x^6}{6} = x^6

step 3

Now, evaluate the antiderivative at the upper limit of integration,

x = 5

:

5^6 = 15625

step 4

Next, evaluate the antiderivative at the lower limit of integration,

x = -2

:

(-2)^6 = 64

step 5

Subtract the value at the lower limit from the value at the upper limit:

15625 - 64 = 15561

Answer

15,561

Key Concept

Definite Integral Evaluation

Explanation

To evaluate a definite integral, find the antiderivative of the integrand, then compute the difference between the values of the antiderivative at the upper and lower limits of integration.

what is a datum in kinematics

Solution by Steps

step 1

In kinematics, a "datum" is a reference point or a fixed starting point used to measure and describe the position, velocity, and acceleration of an object

step 2

The datum is essential for defining the coordinate system and establishing a frame of reference for analyzing motion

step 3

For example, in one-dimensional motion, the datum could be the origin (0,0) of a coordinate system, from which distances and displacements are measured

step 4

In practical applications, the datum helps in setting initial conditions and simplifying the mathematical description of motion

Answer

A datum in kinematics is a reference point used to measure and describe motion.

Key Concept

Datum in Kinematics

Explanation

A datum is a fixed reference point used to define the position, velocity, and acceleration of an object in motion. It helps establish a coordinate system and simplifies the analysis of motion.

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