Equal areas The area of the shaded region under the curve y = 2 sin 2x in (a) equals the area of the shaded region under the curve y = sin x in (b). Explain why this is true without comput- ing areas. YA y y = 2 sin 2x 2 2- y = sin x (a) (b) ele

Q: (1 point) Find the area lying outsider = 2 sin 0 and inside r = 1 + sin 0. Area =

A:

Q: Write the coordinates for the center of the hyperbola: A/ (x-2) (y + 1)² 64 36

A: given  (x-2)236-(y+1)264=1standard form of hyperbola is  (x-h)2a2-(y-k)2b2=1hwer centre is  (h,k)

Q: Vsinx Evaluate dx Vsinx+\cosx

A:

Q: In xdx

A:

Q: 1. 1+tan (0²) = sec (0²)

A:

Q: Evaluate (a) 2. Sinx+ X=X (1) Sinx dx T+x²

A:

Q: integral sec2(?) tan2(?) d?

A: Given that sec2?tan2?d?To find : Integral Solution :∫sec2?tan2?d?Let, tan?=tsec2?d?=dt∫t2dtt33+c

Q: Approximate sinx dx using four trapezoids of equal width. On the interval (0, pi)

A: Trapezoidal rule

Q: Help use change of variables to find image attached

A:

Q: sin 0 do

A:

Q: Diffirentiate 4=V sec? (St?)

A: Write the given expression using exponents, the apply the power rule