Evaluate using Integration by Parts MCQ Quiz in मल्याळम - Objective Question with Answer for Evaluate using Integration by Parts - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Concept: Integration by part

If f and g are two functions, then ∫fg = f∫g - ∫{f'∫g}

If ∫ f(x)dx = F(x), then \(\int_{a}^{b}\)f(x) = F(b)-F(a)

Let \(I=\int_{0}^{π/2}{x^3\sin x}dx\)

We have f = x³ and g = sinx

∴ \(I=x^3\int{\sin x}dx-\int({3x^2\int{\sin x}}dx)dx\)

⇒ \(I=-x^3{\cos x}-\int{-3x^2{\cos x}}dx\)

\(⇒ I=-x^3{\cos x}+3\int{x^2{\cos x}}dx\)

\(⇒ I=-x^3{\cos x}+3\Big[x^2{\sin x}-\int{2x{\sin x}}dx\Big]\)

\(⇒ I=-x^3{\cos x}+3x^2{\sin x}-6\int{x{\sin x}}dx\)

\(⇒ I=-x^3{\cos x}+3x^2{\sin x}-6\Big[-x{\cos x}-\int{-{\cos x}}dx\Big]\)

\(⇒ I=-x^3{\cos x}+3x^2{\sin x}+6x{\cos x}-6\int{{\cos x}}dx\)

\(⇒ I=-x^3{\cos x}+3x^2{\sin x}+6x{\cos x}-6{\sin x}+C\)

\(⇒ I=(3x^2-6){\sin x}+(6x-x^3){\cos x}+C\)

On applying limits of integration,

∴ I = (3(π /2)² - 6)
⇒ \(I=\frac{3\pi^2}{4}-6\)

Concept:

Integration By Parts of Definite Integrals formula: \(\int_{a}^{b}uvdx=u\int_{a}^{b}vdx-\int_{a}^{b}(u' \int vdx)dx\)

Calculation:

Given:

\(I_n=\int^e_1x^{19}(\log|x|)^n dx\)

Here, u=\((\log|x|)^n \) , v = \(x^{19}\)

So apply by Parts:

\(=(\log|x|)^n \int_{1}^{e}x^{19}dx-\int_{1}^{e}(n\frac{(\log|x|)^{n-1}}{x} \int x^{19}dx)dx\) --------(Since u' = \(n\frac{(\log|x|)^{n-1}}{x}\))

\(=(\log|x|)^n \int_{1}^{e}x^{19}dx-\int_{1}^{e}(n\frac{(\log|x|)^{n-1}}{20x} x^{20})dx\) -------( Since \( \int x^{19}dx = x^{20}/20\))

Put the limits we get,

\(I_n = \frac{e^{20}}{20}-\frac{n}{20}(I_{n-1})\)

\(20I_n = {e^{20}}-{n}(I_{n-1})\)

Now,

\(20I_{10} = {e^{20}}-{n}(I_{9})\) ---(1)

\(20I_{9} = {e^{20}}-{n}(I_{8})\) ----(2)

Subtract (1) from (2)

(20)I10 = 10I9 + 9I8

Comparing with (20)I10 = αI9 + βI8

we get

α = 10 and β = 9

So  α - β = 1