Question

Q 36 Determine the value of f(1) + g(1).
Ops: A.
0-2
2
B. C.
Oo
D.
o​

Answer 1

The value of f(1) + g(1) = 1

Correct question : Determine the value of f(1) + g(1) when

$\displaystyle \sf \int {x}^{2} cosx \: dx = f(x) \: sinx + g(x) \: cosx + c$

Given :

$\displaystyle \sf \int {x}^{2} cosx \: dx = f(x) \: sinx + g(x) \: cosx + c$

To find :

The value of f(1) + g(1)

Formula Used :

Integration by Parts

$\displaystyle \sf \int \:f(x)g(x)dx = f(x) \int g(x)dx - \int\bigg[f'(x) \int \: g(x)dx\bigg] dx$

Solution :

Step 1 of 3 :

Integrate the LHS

$\displaystyle \sf \int {x}^{2} cosx \: dx$

$\displaystyle \sf = {x}^{2} \int cosx \: dx - \int \bigg[\frac{d}{dx} ( {x}^{2} )\int cosx \: dx\bigg] dx$

$\displaystyle \sf = {x}^{2} \: sin x - \int 2x \: sinx \: dx$

$\displaystyle \sf = {x}^{2} \: sin x - \bigg[2x\int sinx \: dx - \int \bigg \{ \frac{d}{dx} ( 2x)\int sinx \: dx \bigg \}dx\bigg]$

$\displaystyle \sf = {x}^{2} \: sin x - \bigg[ - 2x \: cosx - \int \bigg \{ - 2cosx \bigg \}dx\bigg]$

$\displaystyle \sf = {x}^{2} \: sin x - \bigg[ - 2x \: cosx + \int 2cosx dx\bigg]$

$\displaystyle \sf = {x}^{2} \: sin x + 2x \: cosx - 2 \int cosx dx$

$\displaystyle \sf = {x}^{2} \: sin x + 2x \: cosx - 2 \: sinx + c$

$\displaystyle \sf = ( {x}^{2} - 2) \: sin x + 2x \: cosx + c$

Step 2 of 3 :

Find f(x) and g(x)

$\displaystyle \sf \int {x}^{2} cosx \: dx = f(x) \: sinx + g(x) \: cosx + c$

$\displaystyle \sf{ \implies }( {x}^{2} - 2) \: sin x + 2x \: cosx + c= f(x) \: sinx + g(x) \: cosx + c$

Comparing both sides we get

$\displaystyle \sf f(x) = {x}^{2} - 2 \: \: \: \: and \: \: \: \: g(x) = 2x$

Step 3 of 3 :

Determine the value of f(1) + g(1)

$\displaystyle \sf f(x) = {x}^{2} - 2 \: \: \: \: and \: \: \: \: g(x) = 2x$

Putting x = 1 we get

$\displaystyle \sf f(1) = {1}^{2} - 2 = 1 - 2 = - 1$

$\displaystyle \sf g(1) = 2 \times 1 = 2$

Thus we get

$\displaystyle \sf f(1) + g(1)$

$\displaystyle \sf = - 1 + 2$

$\displaystyle \sf = 1$

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