Let f0 - 0 and -02f-2tef2tdt-5- then the value of f4 is-

We have nbsp;∫_0^2f'(2t)e^f(2t)dt=5 Put nbsp; e^f(2t)=y⇒ 2f'(2t)e^f(2t)dt=dy Now 12∫_e^f(0)^e^f(4)dy=5⇒∫_e^f(0)^e^f(4)dy=10 ⇒ e^f(4) e^f(0)=10⇒ e^f(4)=10+e ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

Let f0 = 0 an...

Question

Let $f (0) = 0$ and $2 \int 0 f^{'} (2 t) e^{f (2 t)} d t = 5$ , then the value of $f (4)$ is:

2 ln 3

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ln 10

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ln 11

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2 ln 2

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Solution

The correct option is C $ln 11$
We have $2 \int 0 f^{'} (2 t) e^{f (2 t)} d t = 5$
Put $e^{f (2 t)} = y \Rightarrow 2 f^{'} (2 t) e^{f (2 t)} d t = d y$
Now $\frac{1}{2} e^{f (4)} \int e^{f (0)} d y = 5 \Rightarrow e^{f (4)} \int e^{f (0)} d y = 10$
$\Rightarrow e^{f (4)} - e^{f (0)} = 10 \Rightarrow e^{f (4)} = 10 + e^{0} = 11$
Hence $f (4) = ln 11$ .

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Let f(0)=0 and 2∫0f′(2t)ef(2t)dt=5, then the value of f(4) is:

The correct option is C ln11We have 2∫0f′(2t)ef(2t)dt=5 Put ef(2t)=y⇒2f′(2t)ef(2t)dt=dy Now 12ef(4)∫ef(0)dy=5⇒ef(4)∫ef(0)dy=10 ⇒ef(4)−ef(0)=10⇒ef(4)=10+e0=11 Hence f(4)=ln11.

Let $f (0) = 0$ and $2 \int 0 f^{'} (2 t) e^{f (2 t)} d t = 5$ , then the value of $f (4)$ is: