QUESTION 4: INTEGRATION 1 Simplify the following: 2 Integrate the following: 3 Evaluate the following:

Here's the solution to Questions 3 and 4: Question 3: DIFFERENTIATION 3.3.3 Differentiate y = x^2 · 3x (Using product rule). Leave your answers in positive exponents as well as in surd form. Step 1: Identify u and v for the product rule. Let u = x^2 and v = 3x. Step 2: Find the derivatives of u and v. (du)/(dx) = (d)/(dx)(x^2) = 2x (dv)/(dx) = (d)/(dx)( 3x) = (1)/(3x) · (d)/(dx)(3x) = (1)/(3x) · 3 = (1)/(x) Step 3: Apply the product rule formula (dy)/(dx) = u'v + uv'. (dy)/(dx) = (2x)( 3x) + (x^2)((1)/(x)) (dy)/(dx) = 2x 3x + x Step 4: Factor out x for a simplified form. (dy)/(dx) = x(2 3x + 1) The answer is in positive exponents and does not contain surds. 3.4 Determine the co-ordinates of the points of inflection for y = x; 0^ x 270^ with the AID of DIFFERENTIATION. Step 1: Find the first derivative of y = x. (dy)/(dx) = x Step 2: Find the second derivative. (d^2y)/(dx^2) = (d)/(dx)( x) = - x Step 3: Set the second derivative to zero to find potential points of inflection. x = 0 x = 0 For the given interval 0^ x 270^, the values of x for which x = 0 are x = 0^ and x = 180^. Step 4: Check for a change in concavity around these points. • At x = 0^: For x > 0^ (e.g., x=1^), x > 0, so (d^2y)/(dx^2) = - x < 0. The graph is concave down. Since 0^ is an endpoint, concavity change cannot be fully assessed. • At x = 180^: For x < 180^ (e.g., x=170^), x > 0, so (d^2y)/(dx^2) = - x < 0. (Concave down) For x > 180^ (e.g., x=190^), x < 0, so (d^2y)/(dx^2) = - x > 0. (Concave up) Since the concavity changes at x = 180^, this is a point of inflection. Step 5: Find the y-coordinate for the point of inflection. Substitute x = 180^ into the original equation y = x: y = (180^) = 0 The point of inflection is (180^, 0). Question 4: INTEGRATION 4.1 Simplify the following: (sqrt(1 + ^2 x) · x)\,dx Step 1: Use the trigonometric identity 1 + ^2 x = ^2 x. sqrt(1 + ^2 x) = sqrt(^2 x) = | x| Assuming x is in a domain where x > 0, we have | x| = x. Step 2: Substitute the simplified term back into the integral. ( x · x)\,dx Step 3: Integrate the expression. Recall that the derivative of x is - x x. Therefore, the integral of x x is - x. ( x · x)\,dx = - x + C 4.2 Integrate the following: (e^-3x + ( x + 2x)/( x) + sqrt(x) + 2t^3 + (3)/(x) + x)\,dx Step 1: Integrate each term separately. • For e^-3x: e^-3x\,dx = -(1)/(3)e^-3x • For ( x + 2x)/( x): First, simplify the expression: ( x + 2x)/( x) = ( x)/( x) + 2 x x x = ( x)/(^2 x) + (2 x x)/( x) = x x + 2 x Now integrate: ( x x + 2 x)\,dx = x - 2 x • For sqrt(x): sqrt(x)\,dx = x^1/2\,dx = x^1/2+11/2+1 = x^3/23/2 = (2)/(3)x^3/2 • For 2t^3: (assuming t is a constant with respect to x) 2t^3\,dx = 2t^3 x • For (3)/(x): (3)/(x)\,dx = 3 |x| • For x: (assuming is a constant with respect to x) x\,dx = (x^2)/(2) Step 2: Combine all the integrated terms and add the constant of integration C. -(1)/(3)e^-3x + x - 2 x + (2)/(3)x^3/2 + 2t^3 x + 3 |x| + ( x^2)/(2) + C 4.3 Evaluate the following: 4.3.1 _1^2 (3t^2 + 2t)\,dt Step 1: Integrate the function with respect to t. (3t^2 + 2t)\,dt = (3t^3)/(3) + (2t^2)/(2) = t^3 + t^2 Step 2: Evaluate the definite integral using the limits of integration. [t^3 + t^2]_1^2 = (2^3 + 2^2) - (1^3 + 1^2) = (8 + 4) - (1 + 1) = 12 - 2 = 10 4.3.2 _0^/2 sqrt(^2 )\,d Step 1: Simplify the integrand. sqrt(^2 ) = | | For the interval 0 ()/(2), is non-negative. So, | | = . The integral becomes: _0^/2 \,d Step 2: Integrate . \,d = Step 3: Evaluate the definite integral using the limits of integration. [ ]_0^/2 = (()/(2)) - (0) = 1 - 0 = 1 3 done, 2 left today. You're making progress.