Give the calculation in English plz

here is the solution to question 4: a) i) 3x(2-x^2)^(1)/(2) dx; Step 1: Use substitution. Let u = 2-x^2. Then, differentiate u with respect to x: (du)/(dx) = -2x. Rearrange to find dx: dx = (du)/(-2x). Substitute u and dx into the integral: 3x(u)^(1)/(2) ((du)/(-2x)) Step 2: Simplify and integrate with respect to u. The x terms cancel out: -(3)/(2) u^(1)/(2) du Integrate using the power rule u^n du = u^n+1n+1 + C: -(3)/(2) · (u^1)/(2)+1(1)/(2)+1 + C -(3)/(2) · (u^3)/(2)(3)/(2) + C -u^(3)/(2) + C Step 3: Substitute back u = 2-x^2. -(2-x^2)^(3)/(2) + C ii) x^2 3x \, dx; Step 1: Use integration by parts, u \, dv = uv - v \, du. Let u = x^2 and dv = 3x \, dx. Then du = 2x \, dx and v = 3x \, dx = (1)/(3) 3x. x^2 3x \, dx = x^2 ((1)/(3) 3x) - ((1)/(3) 3x) (2x \, dx) = (1)/(3) x^2 3x - (2)/(3) x 3x \, dx Step 2: Apply integration by parts again for x 3x \, dx. Let u_1 = x and dv_1 = 3x \, dx. Then du_1 = dx and v_1 = 3x \, dx = -(1)/(3) 3x. x 3x \, dx = x (-(1)/(3) 3x) - (-(1)/(3) 3x) dx = -(1)/(3) x 3x + (1)/(3) 3x \, dx = -(1)/(3) x 3x + (1)/(3) ((1)/(3) 3x) + C_1 = -(1)/(3) x 3x + (1)/(9) 3x + C_1 Step 3: Substitute the result back into the main integral. (1)/(3) x^2 3x - (2)/(3) (-(1)/(3) x 3x + (1)/(9) 3x) + C = (1)/(3) x^2 3x + (2)/(9) x 3x - (2)/(27) 3x + C The integral is (1)/(3) x^2 3x + (2)/(9) x 3x - (2)/(27) 3x + C. iii) (22-6x)/(x^2+2x-3) dx; Step 1: Factor the denominator and perform partial fraction decomposition. The denominator is x^2+2x-3 = (x+3)(x-1). Set up the partial fractions: (22-6x)/((x+3)(x-1)) = (A)/(x+3) + (B)/(x-1) Multiply both sides by (x+3)(x-1): 22-6x = A(x-1) + B(x+3) To find A, set x=-3: 22-6(-3) = A(-3-1) + B(-3+3) 22+18 = -4A 40 = -4A A = -10 To find B, set x=1: 22-6(1) = A(1-1) + B(1+3) 16 = 4B B = 4 Step 2: Rewrite the integral using partial fractions. ((-10)/(x+3) + (4)/(x-1)) dx Step 3: Integrate each term. -10 (1)/(x+3) dx + 4 (1)/(x-1) dx = -10 |x+3| + 4 |x-1| + C The integral is -10 |x+3| + 4 |x-1| + C. b) To determine the area of the region bounded by the parabola y = x^2+4 and the straight line y = x+6. Step 1: Find the points of intersection of the two curves. Set the equations equal to each other: x^2+4 = x+6 Rearrange into a quadratic equation: x^2 - x - 2 = 0 Factor the quadratic equation: (x-2)(x+1) = 0 The points of intersection are x = -1 and x = 2. These will be the limits of integration. Step 2: Determine which function is above the other in the interval [-1, 2]. Choose a test point within the interval, for example, x=0. For the parabola: y = (0)^2+4 = 4. For the line: y = 0+6 = 6. Since 6 > 4, the line y=x+6 is above the parabola y=x^2+4 in the interval [-1, 2]. Step 3: Set up the definite integral for the area. The area A is given by the integral of the upper function minus the lower function: A = _-1^2 ((x+6) - (x^2+4)) dx A = _-1^2 (-x^2 + x + 2) dx Step 4: Evaluate the definite integral. Integrate term by term: A = [-(x^3)/(3) + (x^2)/(2) + 2x]_-1^2 Apply the limits of integration: A = (-((2)^3)/(3) + ((2)^2)/(2) + 2(2)) - (-((-1)^3)/(3) + ((-1)^2)/(2) + 2(-1)) A = (-(8)/(3) + (4)/(2) + 4) - (-(-1)/(3) + (1)/(2) - 2) A = (-(8)/(3) + 2 + 4) - ((1)/(3) + (1)/(2) - 2) A = (-(8)/(3) + 6) - ((1)/(3) + (1)/(2) - (4)/(2)) A = ((-8+18)/(3)) - ((2+3-12)/(6)) A = (10)/(3) - (-(7)/(6)) A = (10)/(3) + (7)/(6) Find a common denominator: A = (20)/(6) + (7)/(6) A = (27)/(6) Simplify the fraction: A = (9)/(2) The area of the region is (9)/(2) square units. You're doing great! What's next?