integral of dx/x√2x please answer this integral
1. integral of dx/x√2x please answer this integral
∫dx/x√2x
simplify: x√2x =[tex] x^{ \frac{3}{2} } \sqrt{2} [/tex]
∫dx/[tex] x^{ \frac{3}{2} } \sqrt{2} [/tex]
Put the constant outside.
[tex] \frac{1}{ \sqrt{2}} [/tex] ∫dx/[tex] x^{ \frac{3}{2}} [/tex]
Put the x in the numerator the apply power rule.
[tex] \frac{1}{ \sqrt{2} } [/tex] ∫[tex] x^{ -\frac{3}{2} } [/tex]dx
[tex] \frac{1}{ \sqrt{2} } [/tex]([tex]- 2x^{-\frac{1}{2}} [/tex])
[tex] \frac{1}{ \sqrt{2}}( \frac{-2}{\sqrt{x} } } )[/tex]
Answer:
[tex] \frac{-2}{ \sqrt{2x} } [/tex] + C
2. integral of (e^(2x) + 1)^3 e^(2x) dx
Take the integral:
integral e^2 x (e^2 x + 1)^3 dx
Factor out constants:
= e^2 integral x (e^2 x + 1)^3 dx
For the integrand x (e^2 x + 1)^3, substitute u = e^2 x + 1 and du = e^2 dx:
= 1/e^2 integral(u - 1) u^3 du
Expanding the integrand (u - 1) u^3 gives u^4 - u^3:
= 1/e^2 integral(u^4 - u^3) du
Integrate the sum term by term and factor out constants:
= 1/e^2 integral u^4 du - 1/e^2 integral u^3 du
The integral of u^4 is u^5/5:
= u^5/(5 e^2) - 1/e^2 integral u^3 du
The integral of u^3 is u^4/4:
= u^5/(5 e^2) - u^4/(4 e^2) + constant
Substitute back for u = e^2 x + 1:
= (e^2 x + 1)^5/(5 e^2) - (e^2 x + 1)^4/(4 e^2) + constant
Factor the answer a different way:
= ((e^2 x + 1)^4 (4 e^2 x - 1))/(20 e^2) + constant
Which is equivalent for restricted x values to:
Answer: |
| = e^2 ((e^6 x^5)/5 + (3 e^4 x^4)/4 + e^2 x^3 + x^2/2) + constant
3. integration of (2x + 7)dx/x^2 + 2x +5
Answer:
[tex]2(2dx + 7d + 5x)[/tex]
paki check po...4. integrate x^3 e^2x dx
Answer:
Would need to use integration by parts: ∫u dv = uv - ∫v du
∫x3 * e2x dx
where we pick u = x3 and dv = e2x dx
du = d/dx x3= 3x2 dx
v = ∫ dv = ∫ e2x dx = e2x/2
x3 * e2x/2 - ∫ e2x/2 * 3x2 dx
= x3 * e2x/2 - 3/2 ∫ x2e2x dx
Now just need to evaluate ∫ x2e2x dx which requires another integration by parts.
We pick u = x2 and dv = e2x dx.
du = d/dx x2 = 2x dx
v = ∫ dv = ∫ e2x dx = e2x/2
x2 * e2x/2 - ∫ e2x/2 * 2x dx = x2 * e2x/2 - ∫ xe2x dx
Now have to evaluate ∫ xe2x dx which requires one final integration by parts.
Pick u = x and dv = e2x dx.
du = dx
v = ∫ dv = ∫ e2x dx = e2x/2
xe2x/2 - ∫ e2x/2 dx = xe2x/2 - 1/4 e2x
Now we have to go backwards and plug everything that needed to be plugged in with a + C at the end:
x3 * e2x/2 - 3/2 (x2 * e2x/2 - (xe2x/2 - 1/4 e2x)) + C
Feel free to reach out to me if you have questions.
5. What is the largest integral value of x that belongs to the solution of 2x + 12 < 3 ?
Answer:
x = -5
Step-by-step explanation:
— 2x + 12 < 3
— 2x < 3 - 12
— 2x < -9
— x < 9/2: x = -5
6. What is the integral of the function f(x) = sin 2x?
Answer:
2x+ I dont know what is in my head now
7. what are the integral zeros of 3x²+2x-1
Answer:
(x+1)(3x-1)
Step-by-step explanation:
=3x²+3x-x-1
=3x(x+1)(x+1)
=(x+1)(3x-1)
8. sin^(2)(x) csc^(2 )(2x) integrate
Answer:
1/4 tan x + c
Step-by-step explanation:
∫sin²x csc²2x dx = ∫ (sin²x / sin²2x)dx
REMEMBER THE IDENTITY: sin 2x = 2 sinx cosx
∫sin²x csc²2x dx = ∫ [sin²x / ( 2 sinx cosx)² dx]
= ∫ (sin²x / 4 sin²x cos²x) dx
So, sin²x will be cancelled out
= ∫ (1 / 4 cos²x) dx
Factoring out constant
= 1/4 ∫ (1/cos²x) dx
= 1/4 ∫ sec²x dx
= 1/4 tan x + c
9. By integration determine the area bounded by the curves y=6x-x² and y=x²-2x
Answer:
wag Mo nang hanapin si EX
Step-by-step explanation:
that's my best advice for you
10. find the integral ꭍ tan^2 2x sec^4 x dx
di ko po makuha pag pinagsasama lahat.
write in complete details pl
11. integrate sin (2x) ^ 2 * cos(x) dx from 0 to 2pi
Answer:
hi ang alam ko lang is letrastamanlead
Answer:
5.3
Step-by-step explanation:
Since
2(1/3 x3)] 0 to 2
= 16/3
12. integral of x(2x-1)^7 dx
Answer:
18 po ang sagot :)
Step-by-step explanation:
Yan po sagot ko
Answer:
13 sagot po
Step-by-step explanation:
thanks me later
13. Find quadratic equation with the least integral coefficients whose roots are twice that of 2x - 5x + 2 = 0
Answer:
-3x+2=0
Step-by-step explanation:
2x-5x=-3x
-3x+2=0
14. Evaluate the integral of (2x + 3) / x^2 dx.
Answer:
We can rewrite the integral as 2x / x^2 dx + 3 / x^2 dx, and then use the power rule of integration to get (-2/x) - (3/x^2) + C, where C is the constant of integration.
Answer:
yerersss
Step-by-step explanation:
15. What is the largest integral value of x if 3 – 2x > 4?
Answer:
4
Step-by-step explanation:
4 because3-2x =1 the answer is 1
16. Integral of 2x(x+2)³dx
[tex]\large \bold {SOLUTION}[/tex]
[tex]\small\sf{\int{2x(x + 2) ^{3} }\,dx} \\ [/tex]
[tex]\small\textsf{Distribute and evaluate the poynomial}[/tex]
[tex]\small\sf{2\int{x(x + 2) ^{3} }\,dx} \\ [/tex]
[tex]\small\sf{2\int{x(x + 2) ( {x}^{2} + 2x + 2x + 4) }\,dx} \\ [/tex]
[tex]\small\sf{2\int{x(x + 2) ( {x}^{2} + 4x + 4) }\,dx} \\ [/tex]
[tex]\small\sf{2\int{ x({x}^{3} + 2 {x}^{2} + 4 {x}^{2} + 8x + 4x + 8 )}\,dx} \\ [/tex]
[tex]\small\sf{2\int{x( {x}^{3} + 6 {x}^{2} + 12x + 8) }\,dx} \\[/tex]
[tex]\small\sf{2\int{ {x}^{4} + 6 {x}^{3} + 12 {x}^{2} + 8x }\,dx} \\[/tex]
[tex]\small\textsf{By using the Power Rule}[/tex]
[tex]\small\sf{\int{ {x}^{n} } \,dx} = \frac{ {x}^{n + 1} }{n + 1} \\ [/tex]
[tex]\small\sf{\int{ {x}^{4} } \,dx} = \frac{ {x}^{4 + 1} }{4 + 1} \implies \frac{ {x}^{5} }{5} \\ [/tex]
[tex]\small\sf{\int{ {6x}^{3} } \,dx} = \frac{ {6x}^{3 + 1} }{3 + 1} \implies \frac{ {3x}^{4} }{2} \\ [/tex]
[tex]\small\sf{\int{ {12x}^{2} } \,dx} = \frac{ {12x}^{2 + 1} }{2 + 1} \implies {4x}^{3} \\ [/tex]
[tex]\small\sf{\int{ 8x } \,dx} = \frac{ {8x}^{1 + 1} }{1 + 1} \implies 4 {x}^{2} \\ [/tex]
[tex]\small\sf{2( \dfrac{ {x}^{5} }{5} + \dfrac{3 {x}^{4} }{2} + 4 {x}^{3} + 4 {x}^{2}) }[/tex]
[tex]\small\sf{\therefore\int{2x(x + 2) ^{3} }\,dx} = \small\boxed{\green{\sf{ \frac{2 {x}^{5} }{5} + 3 {x}^{4} + 8 {x}^{3} + 8 {x}^{2} +C }}} \\ [/tex]
[tex]\small\textsf{\#CarryOnLearning}[/tex]
[tex] \Large \mathcal{SOLUTION:} [/tex]
[tex] \begin{array}{l} \textsf{Another approach is by u-substitution.} \\ \\ \qquad \large \displaystyle \sf \int 2x(x+2)^3\ dx \\ \\ \textsf{Let u = x + 2} \longrightarrow\sf u - 2 = x \\ \\ \quad \sf du = dx \\ \\ \textsf{The integral becomes} \\ \\ \begin{aligned} \displaystyle \sf \int 2(u - 2)u^3\ du &= \displaystyle \sf 2 \int (u^4 - 2u^3)\ du \\ \\ &= \sf 2 \left(\dfrac{u^5}{5} - \dfrac{2u^4}{4}\right) + C \\ \\ &= \sf \dfrac{2u^5}{5} - u^4 + C \end{aligned} \\ \\ \textsf{Bring back x's} \\ \\ \small \therefore \boxed{\displaystyle \sf \int 2x(x - 2)^3\ dx = \dfrac{2(x + 2)^5}{5} - (x + 2)^4 + C} \end{array} [/tex]
17. integral (3-2x) dx using substitution
Integral (3-2x)dx
Let u = 2x ; du = 2 dx
Integral (3-u)du/2
1/2 integral (3-u)du
1/2 (3u -u^2/2)
3u/2 - u^2 ; u=2x
3(2x)/2 - (2x)^2
3x-4x^2
The answer is 3x-4x^2
18. What is the integral of the function f(x) = sin 2x?
the integral of the function, f(x) = sin 2x is (-1/2) cos (2x) + C
19. what is the integral of cos^2 2x dx
∫cos^2 2x dx
formula:
case III
cos^2 x dx = (1/2 - cos2x/2) dx
=∫(1/2 - cos (2)(2x)/2) dx
=(1/2)∫dx - (1/2)∫cos4x dx
where :
(1/2)(1/4)∫cos4x 4dx
=1/2 x - 1/8 sin4x + C
20. integrate sqrt(2x - 1) dx from 1/2 to 1
Answer:
The answer is in the picture
#CarryOnLearning
21. what positive integral values of x satisfy 2x + 3 <12?
Answer:
x = {4, 3, 2, 1}Interval notation: [1, 9/2) or [1, 4.5)
Step-by-step explanation:
2x + 3 < 12
2x + 3 - 3 < 12 - 3
2x < 9
2x/2 < 9/2
x < 9/2 or 4.5
The positive integers x that satisfies the given inequality:
x = {4, 3, 2, 1} ⇒ integers are not fractions or decimals
Interval notation: [1, 9/2) or [1, 4.5)
where 9/2 or 4.5 are not included, and the smallest positive integer 1 is included.
Check:
When x = 4:
2x + 3 < 12
2(4) + 3 < 12
8 + 3 < 12
11 < 12 (True)
When x = 3:
2(3) + 3 < 12
9 < 12 (True)
22. what is the largest integral value of x if 3 - 2x > 4?
Answer:
4
Step-by-step explanation:
4 because 3-2x = 1 answer is 1
1+1=2 make ur answer arigato
23. laws of positive integral exponents (2x)3
Answer:
[tex]3[/tex]
Arewer of having the selection process
24. express 2x-² as a positive integral exponent
Answer:
2x-²---> [tex]\frac{2}{x^2}[/tex]
Step-by-step explanation:
25. evaluate the integral of ( 4-3sin 2x )⁴ check by differentiation: logarithm
Answer:
3.6 The Chain Rule
Learning Objectives
State the chain rule for the composition of two functions.
Apply the chain rule together with the power rule.
Apply the chain rule and the product/quotient rules correctly in combination when both are necessary.
Recognize the chain rule for a composition of three or more functions.
Describe the proof of the chain rule.
We have seen the techniques for differentiating basic functions (x^n, \, \sin x, \, \cos x, etc.) as well as sums, differences, products, quotients, and constant multiples of these functions. However, these techniques do not allow us to differentiate compositions of functions, such as h(x)= \sin (x^3) or k(x)=\sqrt{3x^2+1}. In this section, we study the rule for finding the derivative of the composition of two or more functions.
Deriving the Chain Rule
When we have a function that is a composition of two or more functions, we could use all of the techniques we have already learned to differentiate it. However, using all of those techniques to break down a function into simpler parts that we are able to differentiate can get cumbersome. Instead, we use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
To put this rule into context, let’s take a look at an example: h(x)= \sin (x^3). We can think of the derivative of this function with respect to x as the rate of change of \sin (x^3) relative to the change in x. Consequently, we want to know how \sin (x^3) changes as x changes. We can think of this event as a chain reaction: As x changes, x^3 changes, which leads to a change in \sin (x^3). This chain reaction gives us hints as to what is involved in computing the derivative of \sin (x^3). First of all, a change in x forcing a change in x^3 suggests that somehow the derivative of x^3 is involved. In addition, the change in x^3 forcing a change in \sin (x^3) suggests that the derivative of \sin (u) with respect to u, where u=x^3, is also part of the final derivative.
We can take a more formal look at the derivative of h(x)= \sin (x^3) by setting up the limit that would give us the derivative at a specific value a in the domain of h(x)= \sin (x^3).
h^{\prime}(a)=\underset{x\to a}{\lim}\frac{\sin (x^3)- \sin (a^3)}{x-a}.
This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the expression x^3-a^3 to obtain
h^{\prime}(a)=\underset{x\to a}{\lim}\frac{\sin (x^3)- \sin (a^3)}{x^3-a^3} \cdot \frac{x^3-a^3}{x-a}.
From the definition of the derivative, we can see that the second factor is the derivative of x^3 at x=a. That is,
\underset{x\to a}{\lim}\frac{x^3-a^3}{x-a}=\frac{d}{dx}(x^3)|_{x=a}=3a^2.
However, it might be a little more challenging to recognize that the first term is also a derivative. We can see this by letting u=x^3 and observing that as x\to a, \, u\to a^3:
\begin{array}{ll} \underset{x\to a}{\lim}\frac{\sin (x^3)- \sin (a^3)}{x^3-a^3} & =\underset{u\to a^3}{\lim}\frac{\sin u- \sin (a^3)}{u-a^3} \\ & =\frac{d}{du}{(\sin u)}|_{u=a^3} \\ & = \cos (a^3) \end{array}
Thus, h^{\prime}(a)= \cos (a^3) \cdot 3a^2.
In other words, if h(x)= \sin (x^3), then h^{\prime}(x)= \cos (x^3) \cdot 3x^2. Thus, if we think of h(x)= \sin (x^3) as the composition (f\circ g)(x)=f(g(x)) where f(x)=\sin x and g(x)=x^3, then the derivative of h(x)= \sin (x^3) is the product of the derivative of g(x)=x^3 and the derivative of the function f(x)= \sin x evaluated at the function g(x)=x^3. At this point, we anticipate that for h(x)= \sin (g(x)), it is quite likely that h^{\prime}(x)= \cos (g(x))g^{\prime}(x). As we determined above, this is the case for h(x)= \sin (x^3).
Now that we have derived a special case of the chain rule, we state the general case and then apply it in a general form to other composite functions. An informal proof is provided at the end of the section.
Step-by-step explanation:
26. Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A What is the value of A^2?
[tex] \huge \mathbb{PROBLEM:} [/tex]
[tex] \begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array} [/tex]
[tex] \huge \mathbb{SOLUTION:} [/tex]
[tex] \!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array} [/tex]
27. f(x)=2-11x+2x² in integres
Your equation: 22√x2+11x+62√=022x2+11x+62=0
Rules for factoring quadratic equation:
1. Find the product of coefficient of x2x2 and constant term
Product = 22√×62√=2422×62=24
2. Try to express "coefficient of x" as sum of two numbers such that product of the numbers is the product of coefficient of x2x2 and constant term
Coefficient of x = 11 = 3 + 8
We specifically chose 3 and 8 because = 8×3=248×3=24 = product of coefficient of x2x2 and constant term
22√x2+(8+3)x+62√⇒(22√x2+8x)+(3x+62√)22x2+(8+3)x+62⇒(22x2+8x)+(3x+62)
Now here is the trick, you have to take common 22√x22x from the first bracket and 33 from the second bracket.
22√x(x+22√)+3(x+22√)⇒(22√x+3)(x+22√)22x(x+22)+3(x+22)⇒(22x+3)(x+22)
Done.
28. integrate sqrt(3-2x)x^2 dx
[tex] \Large \mathbb{SOLUTION:} [/tex]
[tex] \!\begin{array}{l} \sf Evaluate\ \displaystyle \sf \int \sqrt{3 - 2x}\ x^2\ dx \\ \\ \begin{aligned} \sf Let\quad u &=\sf \sqrt{3 - 2x} \\ \sf u^2 &=\sf 3 - 2x \implies 2u\ du = -2\ dx \\ \sf 2x &=\sf 3 - u^2 \quad \quad -u\ du = dx \\ \sf x &=\sf \dfrac{3 - u^2}{2} \end{aligned} \\ \\ \textsf{So the integral becomes} \\ \\ = \displaystyle \sf \int u\left(\dfrac{3 - u^2}{2}\right)^2(-u)\ du \\ \\ = \displaystyle \sf -\dfrac{1}{4} \int u^2(3 - u^2)^2\ du \\ \\ = \displaystyle \sf -\dfrac{1}{4} \int u^2(9 - 6u^2 + u^4)\ du \\ \\ = \displaystyle \sf -\dfrac{1}{4} \int (9u^2 - 6u^4 + u^6)\ du \\ \\ \textsf{By general power rule,} \\ \\ = \displaystyle \sf -\dfrac{1}{4} \left[9\cdot\dfrac{u^{2 + 1}}{2 + 1} - 6\cdot \dfrac{u^{4+1}}{4 + 1} + \dfrac{u^{6 + 1}}{6 + 1}\right] + C \\ \\ = \sf -\dfrac{1}{4} \left(3u^3 - \dfrac{6u^5}{5} + \dfrac{u^7}{7}\right) + C \\ \\ = \sf -\dfrac{u^7}{28} + \dfrac{3u^5}{10} - \dfrac{3u^3}{4} + C \\ \\ \textsf{Substitute back for }\sf u = (3 - 2x)^{\frac{1}{2}} \\ \\ = \boxed{\boxed{\footnotesize \sf -\dfrac{1}{28}(3 - 2x)^{\frac{7}{2}} + \dfrac{3}{10}(3 - 2x)^{\frac{5}{2}} - \dfrac{3}{4}(3 - 2x)^{\frac{3}{2}}+ C }} \end{array} [/tex]
29. simplify 2x-⁴/3x-² Integral Exponent
Answer:
Lol
Step-by-step explanation:
Step 1 i don't know the answers
30. What is the smallest integral value of x which will satisfy 1 +2x >x-2
1 + 2x > x -2
2x - x > -2 -1
x > -3
Since x must be greater than -3, the smallest integer that is larger than -3 is -2.
Therefore the answer is -2.
Check:
1 + 2x > x - 2
1 + 2(-2) > -2 -2
1 + (-4) > -4
-3 > -4 (Correct!)
