Problem 1. The anti-derivative of 1/(1-2x) is 1/2 ln |1-2x| + C. Note the absolute values in the logarithm; without them, you'd be taking a logarithm of a negative number, which is legitimate in complex analysis but not in real analysis. Anyway, the answer thus becomes 1/2 ln 3, which can be rewritten as ln(sqrt(3)).

Problem 2. y^2 exp(y^2 x) and 2xy exp(y^2 x) respectively.

Problem 3. If dv/dx = x+y, then v = 1/2 x^2 + xy + C(y), where C(y) is something which doesn't depend on x. If dv/dy = x-y, then v = xy - 1/2 y^2 + D(x), where D(x) is something which doesn't depend on y. We want both things to be true, so v has to do look like both of these expressions. One solution is v = 1/2 x^2 + xy - 1/2 y^2; you can also add a constant if you wish. (These are the only solutions).

Problem 4. This is a semi-circle. If theta ranged from 0 to 2 pi instead of just to pi, then it would be a full circle, centered at (2,0) and with radius 1.

Problem 5. The ratio test tells you that the series converges if |x-2| is less than one, and diverges if |x-2| is greater than 1. The only possibilities left are when |x-2| are equal to 1, i.e. when x is 1 or 3. But the series becomes 1-1+1-1... in the first case and 1+1+1+... in the second, and neither series converges (e.g. by the zero test). So the series only converges when |x-2| < 1, i.e. when x is strictly between 1 and 3.

Problem 6. Applying the change of variables x = 2y, dx = 2 dy we get

 int_0^8   1/2  dx/(1+x^2)

Problem 7. If we just had to integrate x^2, the answer would be 1/3 = 0.33. But we have this extra term sin(x^2)/300. But this integrand always fluctuates between -1/300 and 1/300, and we're integrating over an interval of length 1, so we know that this part of the integral isn't going to be any bigger than 1/300 in magnitude. Therefore the answer is 0.33 to two decimal places.