The integral of e^x is e^x + C.

The integral of sin(x) is -cos(x) + C.

The integral of sin(x) is -cos(x) + C.

The integral of sin(x) is -cos(x) + C.

The integral is (1/6)x^6 + C.

Diagonal = √(3² + 4² + 12²) = √(9 + 16 + 144) = √169 = 13.

Diagonal = √(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7.

Diagonal = √(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7 units.

As x approaches 0, sin(x) approaches x, thus lim (x -> 0) (x^2/sin(x)) = lim (x -> 0) (x^2/x) = lim (x -> 0) x = 0.

Factoring gives ((x - 1)(x^3 + x^2 + x + 1))/(x - 1). For x ≠ 1, this simplifies to x^3 + x^2 + x + 1. Thus, lim (x -> 1) = 4.

Factoring gives (x - 2)(x + 5)/(x - 2). For x ≠ 2, this simplifies to x + 5. Evaluating at x = 2 gives 7.

The expression is undefined at x=2. The limit does not exist as the function approaches infinity.

The expression can be factored as ((x - 3)(x + 3))/(x - 3). For x ≠ 3, this simplifies to x + 3. Thus, lim (x -> 3) (x + 3) = 6.

The maximum occurs at x = -b/(2a) = -4/(2*-1) = 2. f(2) = -2^2 + 4(2) + 1 = 5.

f'(x) = -3x^2 + 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. f(2) = 5.

Set f'(x) = 0 to find critical points. The local maximum occurs at x = 2. f(2) = 5.

|A| = √(3^2 + (-4)^2) = √(9 + 16) = √25 = 5.

Area = 1/2 * base * height. Max area occurs when height is maximized, thus Area = 1/2 * 10 * 10 = 50.

Area = 1/2 * base * height. Max area occurs when height is maximized, which is 10 units, giving Area = 50.

Area = 1/2 * base * height. Max area occurs when height is maximized at 10 units, giving Area = 50.

For maximum area, the triangle should be equilateral. Area = (sqrt(3)/4) * (10)^2 = 75 cm².

The maximum occurs at t = -b/(2a) = -32/(2*-16) = 1. h(1) = 64.

The maximum occurs at t = -b/(2a) = 64/(2*16) = 2. h(2) = -16(2^2) + 64(2) + 48 = 80.

The function is a downward-opening parabola. The maximum occurs at x = -b/(2a) = -8/(2*-2) = 2. f(2) = -2(2^2) + 8(2) - 3 = 8.

Midpoint = ((2+4)/2, (3+7)/2) = (3, 5).

The vertex gives the minimum at x = 2. f(2) = 4(2^2) - 16(2) + 20 = 4.

The vertex form gives the minimum at x = 2. f(2) = 2.

The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4*2 + 7 = 3.

The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4*2 + 7 = 3.

The minimum occurs at x = 2. f(2) = 2(2^2) - 8(2) + 10 = 6.