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Definite Integration question

2003 · Shift 0 · Q115
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Definite Integration question

2003 · Shift 0 · Q115

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If f(y)=ey,g(y)=y;y>0f\left( y \right) = {e^y},g\left( y \right) = y;y \gt 0f(y)=ey,g(y)=y;y>0 and F(t)=∫0tf(t−y)g(y)dy,F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,}F(t)=0∫t​f(t−y)g(y)dy, then :
  1. A
    F(t)=te−tF\left( t \right) = t{e^{ - t}}F(t)=te−t
  2. B
    F(t)=1t−te−1(1+t)F\left( t \right) = 1t - t{e^{ - 1}}\left( {1 + t} \right)F(t)=1t−te−1(1+t)
  3. C
    F(t)=et−(1+t)F\left( t \right) = {e^t} - \left( {1 + t} \right)F(t)=et−(1+t)
  4. D
    F(t)=tetF\left( t \right) = t{e^t}F(t)=tet.
View written solutionFree

Correct answer: C

  1. Given functions

We have f(y)=ey,g(y)=y,y>0f(y)=e^y, \qquad g(y)=y, \qquad y>0f(y)=ey,g(y)=y,y>0 and F(t)=∫0tf(t−y)g(y) dy.F(t)=\int_0^t f(t-y)g(y)\,dy.F(t)=∫0t​f(t−y)g(y)dy.

  1. Substitute f(t−y)f(t-y)f(t−y) and g(y)g(y)g(y)

Since f(t−y)=et−y,g(y)=y,f(t-y)=e^{t-y}, \qquad g(y)=y,f(t−y)=et−y,g(y)=y, we get F(t)=∫0tet−y y dy.F(t)=\int_0^t e^{t-y}\,y\,dy.F(t)=∫0t​et−yydy.

  1. Factor out ete^tet

Because et−y=ete−ye^{t-y}=e^t e^{-y}et−y=ete−y, F(t)=et∫0tye−y dy.F(t)=e^t\int_0^t y e^{-y}\,dy.F(t)=et∫0t​ye−ydy.

  1. Evaluate ∫ye−ydy\int y e^{-y}dy∫ye−ydy by integration by parts

Let u=y,dv=e−ydy.u=y, \qquad dv=e^{-y}dy.u=y,dv=e−ydy. Then du=dy,v=−e−y.du=dy, \qquad v=-e^{-y}.du=dy,v=−e−y.

So, ∫ye−ydy=uv−∫v du\int y e^{-y}dy = uv-\int v\,du∫ye−ydy=uv−∫vdu =−ye−y+∫e−ydy= -ye^{-y}+\int e^{-y}dy=−ye−y+∫e−ydy =−ye−y−e−y= -ye^{-y}-e^{-y}=−ye−y−e−y =−(y+1)e−y.=-(y+1)e^{-y}.=−(y+1)e−y.

Hence, ∫0tye−ydy=[−(y+1)e−y]0t\int_0^t y e^{-y}dy = \left[-(y+1)e^{-y}\right]_0^t∫0t​ye−ydy=[−(y+1)e−y]0t​ =−(t+1)e−t−(−1)=-(t+1)e^{-t}-\big(-1\big)=−(t+1)e−t−(−1) =1−(t+1)e−t.=1-(t+1)e^{-t}.=1−(t+1)e−t.

  1. Substitute back into F(t)F(t)F(t)

Therefore, F(t)=et(1−(t+1)e−t)F(t)=e^t\left(1-(t+1)e^{-t}\right)F(t)=et(1−(t+1)e−t) =et−(t+1).=e^t-(t+1).=et−(t+1).

So, F(t)=et−(1+t).\boxed{F(t)=e^t-(1+t)}.F(t)=et−(1+t)​.

  1. Check options
  • A: te−tt e^{-t}te−t — incorrect
  • B: does not match — incorrect
  • C: et−(1+t)e^t-(1+t)et−(1+t) — correct
  • D: tett e^ttet — incorrect

Therefore, the correct option is C.\boxed{\text{C}}.C​.

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