Let I- -025e x-x-dx and I be equal to I-n-1e-k-Find n-k -

Since, x [x] is periodic function with period one. ∴ e^ x [x] has period one ∴ I=∫_0^25× 1e^ x [x]dx=25∫_0^1e^ x [x]dx =25∫_0^1e^ x 0dx=25(e ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Conditional Probability

Let I= ∫025...

Question

Let I= $\int_{0}^{25} e^{x - [x]} d x$ and I be equal to $I = (n + 1) (e + k)$ .Find $n + k$ ?

Open in App

Solution

Suggest Corrections

Similar questions

I_{n} = \int_{0}^{1} (x ln x)^{n} d x

I_{4} = k \int_{2}^{1} x^{4} ln x^{3} d x

k

i = \sqrt{- 1} .

\frac{{(- 1 + i \sqrt{3})}^{21}}{(1 - i)^{24}} + \frac{{(1 + i \sqrt{3})}^{21}}{(1 + i)^{24}} = k,

n = [| k |]

| k | .

n + 5 \sum j = 0 (j + 5)^{2} - n + 5 \sum j = 0 (j + 5)

n

r,

(\begin{matrix} n r \end{matrix}) = {\begin{matrix} ^{n} C_{r}, & if n \geq r \geq 0 0, & otherwise \end{matrix}

k

k \sum i = 0 (\begin{matrix} 10 i \end{matrix}) (\begin{matrix} 15 k - i \end{matrix}) + k + 1 \sum i = 0 (\begin{matrix} 12 i \end{matrix}) (\begin{matrix} 13 k + 1 - i \end{matrix})

S_{k} = {lim}_{n \to \infty} \sum_{i = 0}^{n} \frac{1}{{(k + 1)}^{i}}

\sum_{k = 1}^{n} k S_{k}

\int_{0}^{2} [x^{2} - x + 1] d x

[.]

\frac{k - \sqrt{m}}{n}

k + m + n

Join BYJU'S Learning Program