Applications of integrals #5 - Questions and Answers

Question 1 the graph of g consists of two straight lines and a semicircle. use it to evaluate each integral [Image of a graph of y=g(x) that has a positive sloping line, then a semi-circle that drops below the x-axis, followed by a smaller positive sloping line.] a) ∫(from 0 to 10) g(x)dx b) ∫(from 10 to 30) g(x)dx c) ∫(from 0 to 35) g(x)dx Answer we have to evaluate the integral from the given graph we know that integration of the function f(x) is nothing but the area under the curve f(x). a) we have to evaluate, ∫(from 0 to 10) g(x)dx In the graph of g(x) we can see that between x = 0 and x = 10 g(x) is nothing but the right angled triangle with base 10 and height 20 we know that the area of right angled triangle is given by, A = 1/2 * base * height hence we can say that the area of right angled triangle with base 10 and height 20 is given by A = 1/2 * 10 * 20 = 100 Hence we can say that, ∫(from 0 to 10) g(x)dx = area of right angled triangle with base 10 and height 20 ⇒ ∫(from 0 to 10) g(x)dx = 100 b)we have to evaluate, ∫(from 10 to 30) g(x)dx In the graph of g(x) we can see that between x = 10 and x = 30 g(x) is nothing but the semicircle with radius 10 we know that area of a semicircle with radius r is given by, A = 1/2 πr² Hence we can say that area of a semicircle with radius 10 is given by, A = 1/2 π (10)² = 1/2 π * 100 = 50π But we can see that semicircle is below x axis hence we can say that, ∫(from 10 to 30) g(x)dx = -(area of the semicircle with radius 10) ⇒ ∫(from 10 to 30) g(x)dx = −50π c) we have to evaluate, ∫(from 0 to 35) g(x)dx we can say that, ∫(from 0 to 35) g(x)dx = ∫(from 0 to 10) g(x)dx + ∫(from 10 to 30) g(x)dx + ∫(from 30 to 35) g(x)dx we have, ∫(from 0 to 10) g(x)dx = 100, ∫(from 10 to 30) g(x)dx = −50π Hence we can say that, ∫(from 0 to 35) g(x)dx = 100 - 50π + ∫(from 30 to 35) g(x)dx now we will evaluate, ∫(from 30 to 35) g(x)dx In the graph of g(x) we can see that between x = 30 and x = 35 g(x) is nothing but the right angled triangle with base 5 and height 5 we know that the area of right angled triangle is given by, A = 1/2 * base * height hence we can say that the area of right angle triangle with base 5 and height 5 is given by, A = 1/2 * 5 * 5 = 25/2 Hence we can say that, ∫(from 30 to 35) g(x)dx = area of right angled triangle with base 5 and height 5 ⇒ ∫(from 30 to 35) g(x)dx = 25/2 Put this value in equation 1) we can say that, ∫(from 0 to 35) g(x)dx = 100 – 50π + 25/2 ⇒ ∫(from 0 to 35) g(x)dx = 225/2 – 50π Question 2 Find integral. ∫ tan(7x)dx Answer We have to find the integral: ∫ tan(7x) dx We know the integration formula, ∫ tan x dx = ln |sec x| + C ∫ tan ax dx = (1/a) ln |sec ax| + C Where, C is an arbitrary constant. Applying above formula for the given integral, we got ∫ tan(7x) dx = 1/7 ln |sec 7x| + C Hence, value of the integral is 1/7 ln |sec 7x| + C Question 3 Find the antiderivative for each fnction when c equals 0. f(x) = 1/x Answer ∫ f(x) dx = ∫ (1/x) dx = ln|x| + C = ln|x| + 0 = ln|x| Question 4 Find the most general antiderivative or indefinite integral. ∫ (x + √x)dx Answer Step 1 Make two separate integrals ∫ (x + √x)dx = ∫ (x + x^(1/2))dx = ∫ x dx + ∫ x^(1/2) dx Step 2 Integrate each integral. Used integration of xⁿ = x^(n+1)/(n+1) C=constant Answer: ∫ (x + x^(1/2) ) dx = ∫ xdx + ∫x^(1/2) dx = x²/(1+1) + x^(1/2+1)/(1/2 + 1) + C = x²/2 + x^(3/2)/(3/2) + C = x²/2 + (2x^(3/2)/3) + C Question 5 Evaluate the following integral. ∫ x² ln x dx Answer Step 1 To assess the integrals, given ∫ x² ln x dx formula: part-by-part integration ∫ u dv = uv - ∫ v du Step 2 To assess the integrals, given ∫ x² ln x dx Put u = ln x ⇒ du = dx/x and dv = x² dx ⇒ v = x³/3 Therefore, ∫ x² ln x dx = ln x * x³/3 - ∫ (x³/3) * (1/x) dx = x³ ln x / 3 - 1/3 ∫ x² dx = x³ ln x/3 - (1/3)(x³/3) + C = x³ln x/3 - x³/9 + C = x³/9 (3lnx -1) + C = x³/9 (ln (x³/e)) + C Therefore, ∫ x² ln x dx = x³/9 (ln (x³/e)) + C