Mathematics, 04.01.2020 13:31 hilton894

Calculus question

how to do this?

Calculus question how to do this?

Answers: 3

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Answer from: 20alondra04

$\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = 28ln(7) - 24ln(6) - 4$

General Formulas and Concepts:

Algebra I

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$ Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Logarithmic Derivative: $\displaystyle \frac{d}{dx} [lnu] = \frac{u'}{u}$

Integrals

Definite Integrals

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

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

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

Integration by Parts: $\displaystyle \int {u} \, dv = uv - \int {v} \, du$

[IBP] LIPET: Logs, inverses, Polynomials, Exponentials, Trig

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy$

Step 2: Integrate Pt. 1

Identify and find variables for Integration by Parts.

[LIPET] Set: $\displaystyle u = ln(y)$ [LIPET] Set: $\displaystyle dv = \frac{1}{\sqrt{y}} \ dy$ [u] Differentiate [Logarithmic Derivative]: $\displaystyle \frac{du}{dy} = \frac{1}{y}$ [u] Rewrite: $\displaystyle du = \frac{1}{y} \ dy$ [dv] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle v = 2\sqrt{y}$

Step 3: Integrate Pt. 2

[Integral] Integration by Parts: $\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = \bigg[ 2ln(y)\sqrt{y} \bigg] \bigg| \limits^{49}_{36} - \int\limits^{49}_{36} {\frac{2\sqrt{y}}{y}} \, dy$ [Integral] Simplify/Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = \bigg[ 2ln(y)\sqrt{y} \bigg] \bigg| \limits^{49}_{36} - 2\int\limits^{49}_{36} {\frac{1}{\sqrt{y}}} \, dy$ [Integral] Reverse Power Rule: $\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = \bigg[ 2ln(y)\sqrt{y} \bigg] \bigg| \limits^{49}_{36} - 2 \bigg[ 2\sqrt{y} \bigg] \bigg| \limits^{49}_{36}$ Evaluate [Integration Rule - FTC 1]: $\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = 28ln(7) - 24ln(6) - 2(2)$ Simplify: $\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = 28ln(7) - 24ln(6) - 4$

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

Unit: Integration

Book: College Calculus 10e

Answer from: Quest

answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab in the y coordinate

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

x coordinate

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

y coordinate

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

x coordinate

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

y coordinate

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

x coordinate

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

y coordinate

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

x coordinate

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

y coordinate

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

x coordinate

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

y coordinate

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

x coordinate

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

y coordinate

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

x coordinate

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

y coordinate

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

- r3ktforgood ☕

Answer from: Quest

$3^{4x-5}=\left(\dfrac{1}{27}\right)^{2x+10}\\\\3^{4x-5}=\left(\dfrac{1}{3^3}\right)^{2x+10}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\3^{4x-5}=\left(3^{-3}\right)^{2x+10}\qquad\text{use}\ (a^n)^m=a^{n\cdot m}\\\\3^{4x-5}=3^{-3(2x+10)}\qquad\text{use distributive property}\\\\3^{4x-5}=3^{(-3)(2x)+(-3)(10)}\\\\3^{4x-5}=3^{-6x-30}\iff4x-5=-6x-30\qquad\text{add 5 to both sides}\\\\4x=-6x-25\qquad\text{add 6x to both sides}\\\\10x=-25\qquad\text{divide both sides by 10}\\\\\boxed{x=-2.5}$

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