nd the antiderivative x² + 2x -dr. (r² + 1)'

Q: v°dx + x (x + v) dv = 0

A: Let's find.

Q: Evaluate p(x² + y²)5/2 dA, where D = {(x, y) E R² | 0 < y < 1, - V1- y² < a < 0}.

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Q: 5 — 2х 2.x dx 12х — 422 с. V

A: I=∫5-2x12x-4x2dx = ∫5-2x23x-x2dx = ∫523x-x2dx-∫x3x-x2dx

Q: dx xV2ax – x2 ах

A: This is a problem of integration. Based on the general formula we will solve the question.

Q: Evaluate (423 + 3y? + 2z)dV B = {(x, y, z) | 0 < æ < 3,0 < y < 2,0 < z< 2} .

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Q: 1.n f csc n? de y2 - 2y + 3 2.5; Vy - 3y: + 9y – 21 dy

A:

Q: EVALUATE Sx csc(2x) cot(2x) dx A. -z csc (2r) +in esc (2r) - cot (2-)| + C В. B. z csc (2z) + In…

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Q: 6i = (x+i)(y+i)+1

A:

Q: S(8x3 + x-3)dx

A: Solve by integration

Q: Evaluate (3x²y + y)dx + (x³ +x) Select one:

A: Here we have to evaluate ∫(1,2)(4,5)3x2y+ydx+(x3+x)dy ............(1)  Here P(x,y)=3x2y+y and…

Q: C (3y-six) dx + (7x + √y² +1) dy

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Q: 1 1-¹ ( ( 5 + + 2) ³) 3 = Find ac + b. α ~tbeat C

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Q: +2t +3r dr

A: The given differential equation is: rt2+r2+2tdt+t2+3r2dr=0t2+3r2dr+rt2+r2+2tdt=0  Now comparing this…

Q: dx 7. JIVa² – 2x + 5 (hint: let æ² – 2x + 5 = (z – x)²) -

A: Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: When x = 2s+7t, y=5st and z= 3.x2 – 3, in the form of s and t. express

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Q: Evaluate. 17x - 19 √(x + 5) (22² + 1) dx = (Use c for the constant of antidifferentiation)

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Q: x¹ + x² + 1 I(x² + 1)² -dx.

A: The given integral is: ∫x4+x2+1xx2+12dx Evaluate the given integral.

Q: Evaluate : ((x + Vx² +1) dx.

A: To find ∫x+x2+1ndx

Q: S4x(5x³ – 3x)dx

A: To find the antiderivative: ∫4x5x3-3xdx

Q: e2" – e5r dr dx b)] Jx (1 + x) + e3r

A:

Q: 1 ·5² 6² (+2²³ +²² dx dy. (1 0 0 3. Compute

A: 3. Given that the double integral is                          ∫0∞∫0∞1(1+x2+y2)2dxdy.

Q: Let 7(t) = 2t î + 4t² j. Compute Ñ(1).

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Q: p6x² – x – 14 7. J4 x² – 2x – 3 -

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Q: 5. Save for x T 6. Salve for y. (25y-5 (20x)

A: