Balbharati Maharashtra State BoardStd 12 Commerce Statistics Part 1 Digest PdfChapter 5 Integration Ex 5.1 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.1
Question 1
Maharashtra Board Solution
Evaluate
Solution & Step-by-Step Answer:
Question 2
Maharashtra Board Solution
Evaluate
Solution & Step-by-Step Answer:
Question 3
Maharashtra Board Solution
Evaluate
Solution & Step-by-Step Answer:
Question 4
Maharashtra Board Solution
Evaluate ∫(3x2 – 5)2 dx
Solution & Step-by-Step Answer:
∫(3x2 – 5)2 dx = ∫(9x4 – 30x2 + 25) dx = 9∫x4 dx – 30∫x2 dx + 25∫1 dx = 9() – 30() + 25x + c = – 10x3 + 25x + c.
Question 5
Maharashtra Board Solution
Evaluate
Solution & Step-by-Step Answer:
Question 6
Maharashtra Board Solution
If f'(x) = x2 + 5 and f(0) = -1, then find the value of f(x).
Solution & Step-by-Step Answer:
By the definition of integral f(x) = ∫f'(x) dx = ∫(x2 + 5) dx = ∫x2 dx + 5∫1 dx = + 5x + c Now, f(0) = -1 gives f(0) = 0 + 0 + c = -1 ∴ c = -1 ∴ from (1), f(x) = + 5x – 1.
Question 7
Maharashtra Board Solution
If f(x) = 4x3 – 3x2 + 2x + k, f(0) = -1 and f(1) = 4, find f(x).
Solution & Step-by-Step Answer:
By the definition of integral f(x) = ∫f'(x) dx = ∫(4x3 – 3x2 + 2x + k) dx = 4∫x3 dx – 3∫x2 dx + 2∫x dx + k∫1 dx = 4() – 3() + 2() + kx + c ∴ f(x) = x4 – x3 + x2 + kx + c Now, f(0) = 1 gives f(0) = 0 – 0 + 0 + 0 + c = 1 ∴ c = 1 ∴ from (1), f(x) = x4 – x3 + x2 + kx + 1 Further f(1) = 4 gives f(1) = 1 – 1 + 1 + k + 1 = 4 ∴ k = 2 ∴ from (2), f(x) = x4 – x3 + x2 + 2x + 1.
Question 8
Maharashtra Board Solution
If f(x) = – kx + 1, f(0) = 2 and f(3) = 5, find f(x).
Solution & Step-by-Step Answer:
By the definition of integral
