"Let $F:[3,5] \to R$ be a twice differentiable function on (3, 5) such that $F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt}$. If $F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$, then $\alpha$ + $\beta$ is equal to _______________."

Question

Accepted Answer

"$F(3) = 0$

${e^x}F(x) = \int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt}$

${e^x}F(x) + {e^x}F'(x) = 3{x^2} + 2x + 4F'(x)$

$({e^x} - 4){{dy} \over {dx}} + {e^x}y = (3{x^2} + 2x)$

$${{dy} \over {dx}} + {{{e^x}} \over {({e^x} - 4)}}y = {{(3{x^2} + 2x)} \over {({e^x} - 4)}}$$

$$y{e^{\int {{{{e^x}} \over {({e^x} - 4)}}dx} }} = \int {{{(3{x^2} + 2x)} \over {({e^x} - 4)}}{e^{\int {{{{e^x}} \over {{e^x} - 4}}dx} }}dx} $$

$y.({e^x} - 4) = \int {(3{x^2} + 2x)dx + c}$

$y({e^x} - 4) = {x^3} + {x^2} + c$

Put x = 3 $\Rightarrow$ c = $-$36

$F(x) = {{({x^3} + {x^2} - 36)} \over {({e^x} - 4)}}$

$$F'(x) = {{(3{x^2} + 2x)({e^x} - 4) - ({x^3} + {x^2} - 36){e^x}} \over {{{({e^x} - 4)}^2}}}$$

Now, put value of x = 4 we will get $\alpha$ = 12 & $\beta$ = 4"

Let $F:[3,5] \to R$ be a twice differentiable function on (3, 5) such that

$F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt}$. If $F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$, then $\alpha$ + $\beta$ is equal to _______________.

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Let $F:[3,5] \to R$ be a twice differentiable function on (3, 5) such that $F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt}$. If $F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$, then $\alpha$ + $\beta$ is equal to _______________.

Solution

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More Definite Integration questions

Drill 25 more like these. Every day. Free.

Let $F:[3,5] \to R$ be a twice differentiable function on (3, 5) such that

$F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt}$. If $F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$, then $\alpha$ + $\beta$ is equal to _______________.