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: -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: dx (9 – 4.x²)³/2 2,3/2

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

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

A:

Q: Vx³ Inx dx In x dx 1

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

A:

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