Given e-2-72- e2-7-39- e3-20-09- e4-54-60 the approximate value of -04exdx using Simpson-s rule and taking h-1 is-

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Standard IX

Mathematics

Laws of Exponents for Real Numbers

Given e=2.7...

Question

Given $e = 2.72, e^{2} = 7.39, e^{3} = 20.09$ , $e^{4} = 54.60$ the approximate value of $\int_{0}^{4} e^{x} d x$ using Simpson`s rule and taking $h = 1$ is?

57.325

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53.873

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58.873

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54.525

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Solution

The correct option is A $53.873$
Let $f (x) = e^{x}$
$f (0) = 1 = y_{0}$
$f (1) = 2.72 = y_{1}$
$f (2) = 7.39 = y_{2}$
$f (3) = 20.09 = y_{3}$
$f (4) = 54.60 = y_{4}$
Here $h = 1$ .
Applying Simpson's rule, we get
$\int_{0}^{4} e^{x} . d x$
$= \frac{h}{3} [y_{0} + 2 [y_{2}] + 4 [y_{1} + y_{3}] + y_{4}]$
$= \frac{1}{3} [1 + 2 (7.39) + 4 (2.72 + 20.09) + 54.60]$
$= \frac{1}{3} [161.62]$
$= 53.8733$

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Similar questions

Q. Assertion(

A

): lf

e = 2.72, e^{2} = 7.39

e^{3} = 20.09, e^{4} = 54.6

then

\int_{0}^{4} e^{x} d x

by Simpson's rule is

53.873

Reason(R): Simpson's rule is

\int_{a}^{b} y d x = \frac{h}{2}

[(y_{0} + y_{n}) + 2 (y_{1} + y_{2} + \dots . . + y_{n - 1})]

Q. The ratio of

(E_{2} - E_{1})

(E_{4} - E_{3})

for

H

- atom is approximately.

Q. If

E = E_{1} \cap E_{2} \cap E_{3} \cap E_{4} \cap E_{5}

and

P (E_{1}) = \frac{95}{100}, P (E_{2} / E_{1}) = \frac{94}{99}, P (E_{3} / (E_{1} \cap E_{2})) = \frac{93}{98}, P (E_{4} / E_{1} \cap E_{2} \cap E_{3}) = \frac{92}{97}

and

P (E_{5} / E_{1} \cap E_{2} \cap E_{3} \cap E_{4}) = \frac{91}{96}

, then

P (E)

is:

Q. If

E = E_{1} E_{2} E_{3} E_{4} E_{5}

and

P (E_{1}) = \frac{95}{100}

P (E_{2} / E_{1}) = \frac{94}{99}

P (E_{1} / E_{1} E_{2}) = \frac{93}{98}

P (E_{4} / E_{1} E_{2} E_{3}) = \frac{92}{97}

and

P (E_{5} / E_{1} E_{2} E_{3} E_{4}) = \frac{91}{96}

, then find

P (E)

The figure gives the electric potential $V$ as a function of distance through five regions on $x -$ axis. Which of the following is true for the electric field $E$ in these regions?

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Given e=2.72, e2=7.39, e3=20.09,e4=54.60 the approximate value of ∫40exdx using Simpson`s rule and taking h=1 is?

The correct option is A 53.873Let f(x)=exf(0)=1=y0f(1)=2.72=y1f(2)=7.39=y2f(3)=20.09=y3f(4)=54.60=y4Here h=1.Applying Simpson's rule, we get ∫40ex.dx=h3[y0+2[y2]+4[y1+y3]+y4]=13[1+2(7.39)+4(2.72+20.09)+54.60]=13[161.62]=53.8733

Given $e = 2.72, e^{2} = 7.39, e^{3} = 20.09$ , $e^{4} = 54.60$ the approximate value of $\int_{0}^{4} e^{x} d x$ using Simpson`s rule and taking $h = 1$ is?