University:
Boston University
Course:
CGS MA 121 | Calculus
Academic year:

2024
Views:

233

Pages:

5
Author:

fiziokrat16l

Question 1 Evaluate the integral. ∫(x⁶-7x)⁴ dx Answer 1 Step 1 Consider the integral ∫ (x⁶-7x)⁴ dx Step 2 let I = ∫(x⁶-7x)⁴ dx remove x from the bracket I = ∫ x⁴(x⁵-7)⁴ dx put u = x⁵ - 7 differentiating it wrt x we have du = 5x⁴ dx du/5 = x⁴ dx Replace it with the integral I = ∫ (u⁴/5) du I = 1/5 ∫u⁴du I = (1/5) (u⁵/5) + c I = (x⁵-7)⁵ / 25 + c Answer 2 ∫(x⁶-7x)⁴dx = ∫ x²⁴ - 28x¹⁹ + 294x¹⁴ - 1372x⁹ + 2401x⁴dx = ∫ x²⁴dx - 28∫ x¹⁹dx + 294∫ x¹⁴dx - 1372∫ x⁹dx + 2401∫ x⁴dx = x²⁵/25 - 28x²⁰/5 + 294x¹⁵/15 - 1372x¹⁰/10 + 2401x⁵/5 = x²⁵/25 - 7x²⁰/5 + 98x¹⁵/5 - 686x¹⁰/5 + 2401x⁵/5 + C = x²⁵/25 - 7x²⁰ + 98x¹⁵ - 686x¹⁰ + 2401x⁵/5 + C Answer 3 ∫(x⁶-7x)⁴dx = ∫x⁴(x⁵-7)⁴dx = 1/5 ∫u⁴du where u = x⁵-7 = ∫(u-7)⁴du = ∫v⁴dv = v⁵/5 = (u-7)⁵/5 = 1/25 (x⁵-7)⁵ + C Question 2 Show that the differential forms in the integrals are exact. Then evaluate the integrals. ∫(from 1,1,2 to 3,5,0) yzdx + xzdy + xydz Answer 1 Step 1 To ascertain: demonstrate the accuracy of the integrals' differential forms. The integrals are then evaluated. Given: We possess a fundamental ∫(from 1,1,2 to 3,5,0) yz dx + xz dy + xy dz Explanation:let M = yz, N = xz, P = xy and apply the Test for exactness ∂P/∂y = x = ∂N/∂z ∂M/∂z = y = ∂P/∂x ∂N/∂x = z = ∂M/∂y It is clear from this that the supplied differential form is accurate. Now, let's think about it df = yz dx + xz dy + xy dz for some f, and the integral value is f(3,5,0) -f(1,1,2) Step 2 ∫(from 1,1,2 to 3,5,0) yz dx + xz dy + xy dz = [xyz](from 1,1,2 to 3,5,0) = [3xyz](from 1,1,2 to 3,5,0) = 3[(3×5×0) - (1×1×2)] = -6 Question 3 Evaluate the integral. ∫ cos²(4x)dx Answer Step 1 We use the double angle formula cos (8x) = 2cos²(4x) - 1 2cos²(4x) = cos (8x) +1 cos²(4x) = (cos (8x) +1)/2 Step 2 Then we use that in the given integral ∫ cos²(4x)dx = ∫ (cos (8x) + 1)/2 dx = 1/2 ∫ (cos (8x) + 1) dx = 1/2 [∫ cos (8x)dx + ∫1dx] = 1/2 [sin (8x)/8 + x] + C [since ∫cos(mx)dx = sin(mx)/m] = sin(8x)/16 + x/2 + C Answer: ∫ cos²(4x)dx = sin(8x)/16 + x/2 + C C=integrating constant Question 4 Find the indefinite integral ∫ sec(4x) dx Answer Step 1 Having said that, ∫ sin(t)dt = -cos(t) + C...(1) Given integral is what we have as I = ∫ sin (4x) dx Step 2 Let us consider that, 4x=t 4dx=dt dx = 1/4 dt Thus, integral becomes as I = ∫ sin(t) (1/4) dt = 1/4 ∫ sin(t) dt = 1/4 (-cos(t) + C) (By using equation (1)) = -1/4 cos(t) + C On substituting back t=4x, we get the value of required integral as ∫ sin (4x)dx = -1/4 cos(4x) + C Hence, value of integral ∫ sin (4x)dx is -1/4 cos(4t) + C. Question 5 Find the general indefinite integral. (Use C for the constant of integration.) ∫ (u⁶ - 2u⁵ - u³ + 4/3) du Answer 1 Step 1 Find the general indefinite integral. Given: ∫ (u⁶ - 2u⁵ - u³ + 4/3) du Apply the power rule, ∫ xⁿ dx = x^(n+1)/(n+1) ,n≠-1 Integral of constant, ∫ adx = ax Step 2 Now integrate the term, we get ∫ (u⁶ - 2u⁵ - u³ + 4/3) du = u⁷/7 - 2u⁶/6 - u⁴/4 + 4u/3 + C = u⁷/7 - u⁶/3 - u⁴/4 + 4u/3 + C Answer 2 Step 1: Expand. ∫ (u⁶ - 2u⁵ - u³ + 4/3) du Step 2: Use Power Rule: ∫xⁿdx = x^(n+1)/(n+1) + C. = u⁷/7 - u⁶/3 - u⁴/4 + 4u/3 + C Step 3: Add constant. = u⁷/7 - u⁶/3 - u⁴/4 + 4u/3 + C