∫_0^1 (x^2 + 1) dx = [x^3/3 + x] from 0 to 1 = (1/3 + 1) = 4/3.

The integral evaluates to [x^5/5 + (1/2)x^4 + (1/3)x^3] from 0 to 1 = 1/5 + 1/2 + 1/3 = 31/30.

∫_0^1 (x^4 + 2x^3) dx = [x^5/5 + (1/2)x^4] from 0 to 1 = (1/5 + 1/2) = 7/10.

∫_0^1 x^4 dx = [x^5/5] from 0 to 1 = 1/5.

Using the identity sin(2x) = 2sin(x)cos(x), the integral becomes (1/2)∫_0^π sin(2x) dx = 0.

∫_0^π sin(x) dx = [-cos(x)] from 0 to π = -(-1 - 1) = 2.

The integral evaluates to [x/2 + sin(2x)/4] from 0 to π/2 = π/4.

Solving gives cos^2(x) = 1/3, so x = π/3, 2π/3, and their equivalents.

The solution is x = π/6 + 2nπ.

Factoring gives (sin(x) - 2)(sin(x) + 1) = 0, so sin(x) = 2 (not possible) or sin(x) = -1, giving x = 3π/2.

Factoring gives sin(x)(sin(x) - 1) = 0, so x = 0 and x = π.

Rearranging gives sin(x) = 1/3, which has solutions in the specified interval.

The solution is x = π/2.

Factoring gives sin(x)(sin(x) - 1) = 0, so x = 0 and x = π.

Volume = (4/3)πr³ = (4/3)π(7)³ ≈ 143.67 cm³

Weighted mean = (10*1 + 20*2 + 30*3) / (1 + 2 + 3) = (10 + 40 + 90) / 6 = 140 / 6 = 23.33.

Sad is an antonym of the other three words.

'Fearless' is a synonym for 'Brave'.

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f''(1) < 0 indicates a local maximum at x = 1.

The vertex occurs at x = -b/(2a) = 4/2 = 2, which is the x-coordinate of the minimum point.

The vertex occurs at x = -b/(2a) = 4/2 = 2. This is where the local minimum occurs.

Setting y = 0 in the equation gives 5x + 10 = 0, thus x = -2. The x-intercept is -2.

Setting x = 0 in the equation gives y = 8. Therefore, the y-intercept is 8.

Set x = 0: -2y = 10 => y = -5. The y-intercept is (0, -5).

The integral is (5/5)x^5 + C = x^5 + C.

The integral is (6/3)x^3 - 4x + C = 2x^3 - 4x + C.

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

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

The arrangement is C > A > B and E > D. The middle book is A.

The arrangement is C, A, B, D. Therefore, Book C is positioned leftmost.