Hello Calculus!

Find the value

[tex]\\ \rm\Rrightarrow {\displaystyle{\int\limits_3^5}}(e^{3x}+7cosx-3tan^3x)dx[/tex]

Note:-

Answer with proper explanation required and all steps to be mentioned .

Answers

Answer 1

Answer:

[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]

General Formulas and Concepts:
Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]

Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Integration Method: U-Substitution + U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx[/tex]

Step 2: Integrate Pt. 1

[Integral] Rewrite [Integration Rule - Addition/Subtraction]:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + \int\limits^5_3 {7 \cos x} \, dx - \int\limits^5_3 {3 \tan^3 x} \, dx[/tex][Integrals] Rewrite [Integration Property - Multiplied Constant]:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^3 x} \, dx[/tex][3rd Integral] Rewrite:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution and u-solve.

1st Integral

Set u:
[tex]\displaystyle u = 3x[/tex][u] Differentiate [Derivative Properties and Rules]:
[tex]\displaystyle du = 3 \, dx[/tex][Bounds] Swap:
[tex]\displaystyle \left \{ {{x = 5 \rightarrow u = 3(5) = 15} \atop {x = 3 \rightarrow u = 3(3) = 9}} \right.[/tex]

3rd Integral

Set v:
[tex]\displaystyle v = \sec x[/tex][v] Differentiate [Trigonometric Differentiation]:
[tex]\displaystyle dv = \sec x \tan x \, dx[/tex][dv] Rewrite:
[tex]\displaystyle dx = \frac{1}{\sec x \tan x} \, dv[/tex]

Step 4: Integrate Pt. 3

Let's focus on the 3rd integral first.

Apply Integration Method [U-Solve]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \int\limits^{x = 5}_{x = 3} {\frac{v^2 - 1}{v}} \, dv[/tex][Integral] Rewrite [Integration Property - Addition/Subtraction]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \int\limits^{x = 5}_{x = 3} {v} \, dv - \int\limits^{x = 5}_{x = 3} {\frac{1}{v}} \, dv \Bigg)[/tex][Integrals] Apply Integration Rules [Reverse Power Rule and Logarithmic Integration]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{v^2}{2} \bigg| \limits^{x = 5}_{x = 3} - \ln | v | \bigg| \limits^{x = 5}_{x = 3} \Bigg)[/tex][v] Back-Substitute:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 x}{2} \bigg| \limits^{5}_{3} - \ln | \sec x | \bigg| \limits^{5}_{3} \Bigg)[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 5 - \sec^2 3}{2}- \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \Bigg)[/tex]

Step 5: Integrate Pt. 4

Focus on the other 2 integrals and solve using integration techniques listed above.

1st Integral:

[tex]\displaystyle\begin{aligned}\int\limits^5_3 {e^{3x}} \, dx & = \frac{1}{3} \int\limits^5_3 {3e^{3x}} \, dx \\& = \frac{1}{3} \int\limits^{15}_9 {e^{u}} \, du \\& = \frac{1}{3} e^u \bigg| \limits^{15}_9 \\& = \frac{1}{3} \bigg( e^{15} - e^9 \bigg)\end{aligned}[/tex]

2nd Integral:
[tex]\displaystyle\begin{aligned}7 \int\limits^5_3 {\cos x} \, dx & = 7 \sin x \bigg| \limits^5_3 \\& = 7 \bigg( \sin 5 - \sin 3 \bigg)\end{aligned}[/tex]

Step 6: Integrate Pt. 5

[Integrals] Substitute in integrals:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]

∴ we have evaluated the integral.

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Answer 2

Answer:

1086950.36760

Formula's used:

[tex]\rightarrow \sf \int sin(ax+b)=-\dfrac{1}{a} cos(ax+b)+c[/tex]

[tex]\rightarrow \sf \int cos(ax+b)=\dfrac{1}{a} sin(ax+b)+c[/tex]

[tex]\rightarrow \sf \int \dfrac{1}{ax+b} =\dfrac{1}{a} ln|ax+b|+c[/tex]

[tex]\rightarrow \sf \int e^{ax+b}=\dfrac{1}{a} e^{ax+b} + c[/tex]

[tex]\rightarrow \bold{ ln|a| - ln|b| = ln|\frac{a}{b} | }[/tex]

Explanation:

[tex]\dashrightarrow \sf \int \left(e^{3x}+7cos\left(x\right)-3tan^3\left(x\right)\right)[/tex]

                        apply sum rule: [tex]\bold{\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}[/tex]

[tex]\dashrightarrow \sf \int \:e^{3x}dx+\int \:7\cos \left(x\right)dx-\int \:3\tan ^3\left(x\right)dx[/tex]

                Integrate simple followings first, using formula's given above

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-\int 3tan^3x[/tex]

                        Breakdown the component

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int tan^2x(tanx)[/tex]

                                                         [ tan²x = sec²x - 1 ]

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int (sec^2x-1)(tanx)[/tex]

===========================================================

for integration of [tex]\bold{\int (sec^2x-1)(tanx)}[/tex]

                                                  apply substitution ... u

[tex]\dashrightarrow \int \dfrac{-1+u^2}{u}[/tex]

[tex]\dashrightarrow \sf \int \:-\dfrac{1}{u}+udu[/tex]

[tex]\dashrightarrow \sf - \int \dfrac{1}{u}du+\int \:udu[/tex]

[tex]\dashrightarrow -\ln \left|u\right|+\dfrac{u^2}{2}[/tex]

substitute back u = sec(x)

[tex]\dashrightarrow \sf-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}[/tex]

================================================= insert back

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\left(-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}\right)[/tex]   outcome after integrating

Now apply the given limits

[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{3(5)}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{3(3)}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]

                                                                   simplify

[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{15}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{9}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]

                          and group the variables

[tex]\sf \hookrightarrow \dfrac{e^{15}-e^9}{3}-\dfrac{3}{2\cos ^2\left(5\right)}+\dfrac{3}{2\cos ^2\left(3\right)}+7\sin \left(5\right)-7\sin \left(3\right)+3\ln \left(\dfrac{1}{\cos \left(5\right)}\right)-3\ln \left(-\dfrac{1}{\cos \left(3\right)}\right)[/tex]

value:

[tex]\sf \hookrightarrow 1086950.36760[/tex]


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Answers

Answer:

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Step-by-step explanation:

Using the formula

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Answer:

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Step-by-step explanation:

y x 5 =8

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Answers

Susan's monthly payment will be $117.93.

We have Susan take out a personal loan for $3,500 at an interest rate of 13% compounded monthly.

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A

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888 + 88 + 8 + 8 + 8 = 1000

have a nice day!

888 + 88 + 8 + 8 + 8 = 1000

hope this helps....

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Answer:

D) AD ≅ AE

Step-by-step explanation:

According to given details the triangles ADB and BCE are congruent as per HL congruence

Lets verify the choices

∠DAB ≅ ∠CBA,

this is true as corresponding angles of congruent triangles

AE ≅ BE,

this is true as AEB is isosceles

AC ≅ BD,

this is true as corresponding sides of congruent triangles

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this is NOT true as AE > AD as hypotenuse of right triangle is greater than legs

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this is true as AEB is isosceles

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D) AD ≅ AE

Step-by-step explanation:

Hook me up with a 5 star and a Thanks :)


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Answer:

500

Remember the rounding rule for the ones place...

0 to 4? Round down or keep the same.

5 to 9? Round up.

__________________________________________________________

Questions?

Ask in comments.

It would stay 500! Hope this helped

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16
Red
Blue
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3
Based on these results, express the probability that the next spin will land on
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Answers

Answer:

15.09%

Step-by-step explanation:

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53

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plan c. is the answer byebye

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Answer:

I hope I answered your question!

Step-by-step explanation:

6.80+0.90x=7.30+0.65x (multiply by 100 on both sides)

680+90x=730+65x

50=25x

x=2

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C

Step-by-step explanation:

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40 cm.

Step-by-step explanation:

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Which situation below has the highest constant of
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B) 32 ounces of white rice costs $6.40
C) 12 ounces of white rice costs $1.20.
D) 9 ounces of brown rice costs $1.80

Answers

Answer:

A) k = 0.3

Step-by-step explanation:

Find the constant of proportionality k in each case and compare

y = kx

A)

16k = 4.80 ⇒ k = 4.80/16 ⇒ k = 0.3

B)

32k = 6.40 ⇒ k = 6.40/32 ⇒ k = 0.2

C)

12k = 1.20 ⇒ k = 1.20/12 ⇒ k = 0.1

D)

9k = 1.80 ⇒ k = 1.80/9 ⇒ k = 0.2

As we see k = 0.3 is the highest one

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#A

4.80/160.3

#B

6.4/320.2

#C

1.2/120.1

#D

1.8/90.2

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