Set up an integral for the length of the curve. y = Vi-x7, -sxs 1/4 4+ 49, 12 dx -1/4 4- 4x7 + 7x6 dx 4(1 – x7) 1/4 1/4 1/4 4- 4x7. 1912 dx 4(1 – x7) 1/4 1/4 5- 4x7 dx 1/4 V4(1 -x7)

Q: /3 2 %3D

A: To find: The value of dydx. Given: The equation is y=x35x2+1-23. Formula used: Chain rule of…

Q: dx xーメ

A: Explanation: Given that,  ∫x4+1x3-xdx Now we will perform long division and we will get…

Q: -9 dx (3x + 1)² 1

A: Solution: ∫1∞-9(3x+1)2dxTake the constant out=-9×∫1∞1(3x+1)2dx

Q: dy Find for y= x(7x+9) dx dy dx

A: We use product rule

Q: Integrate .5 X 12 1+x

A: Integration by substitutions

Q: 1 dx %3D x-3V2x-4+2

A:

Q: dx (9 – 4.x²)³/2 2,3/2

A: Given: The integral is ∫dx9-4x232. To find: Value of the integral.

Q: Fnd -13 2/3 dx

A:

Q: ©Use TMe Phopentos to Eselaate dx=D4 (2) 3 x da= 2

A:

Q: -3/2 1/2V2 dx 2 dx 5. 6. V9 - x VI - 4x

A: Integration of the functions and substitution of the limits helps to find the solution of the…

Q: Vx³ Inx dx In x dx 1

A:

Q: 3. X+1 (tah dx X-1

A:

Q: d x – 2x d -{2x° + x=°] dx 5. dx 6. 7. d 5x-1 dx

A:

Q: mark all the points on the plot below where dy/dx=-2

A:

Q: Evaluate ¹/2 1 3 — 1 dx.

A:

Q: 9. x² dx (1 – x²)3/2

A: Let's find.

Q: dy + 2xy = 1. dx

A:

Q: 8x 5. S dx 2. 2. (4x +3)

A: ∫8x(4x2+3)23dx∫(4x2+3)-23 8x dxhere f(x) =4x2+3 , f'(x) =8x , n=-23∫f(x)nf'(x) dx= f(x)n+1n+1+c

Q: -3/2 dx S(35 + 4x – 4x²)

A: Explained below

Q: Væ – 4/x dx

A:

Q: dx 1– ex

A: Consider the following integral: ∫ex1-exdx Substitute u=ex2→du=12ex2dx in the above equation:…

Q: x³ +4x² + x – 1 dx 2 X 16. .3 2 1 x' + x'

A:

Q: - Trig Sub 1 dx V9 – x²

A:

Q: V3 1-x dx V1-x2 0.

A: We can find the solution as below

Q: (2x+7)2 x 2 dx

A: on diffrentiating this with respect to x we get =2(2x+7)ddx(2x+7)=2(2x+7)×2=4(2x+7)

Q: dr Find if r (x) = 9x 2 + 5x – 3. dx

A: Given,            r(x) = 9x2 + 5x - 3

Q: 1 (x2 - y2) 3 dx

A: We have to evaluate ∫1x2-y232dx Let x=y secu Then: dxdu=dduy secu⇒dxdu=y secutanu⇒dx=y secutanu du…