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# Integral

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$$\int_{3}^{4} (x^2-1)*e^x/e-7ex+11e dx=?$$

Whitespy001  Jun 2, 2017
edited by Whitespy001  Jun 2, 2017
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#1
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Compute the definite integral:
integral_3^4 (e^(x - 1) (x^2 - 1) - 7 e x + 11 e) dx

Integrate the sum term by term and factor out constants:
= integral_3^4 e^(x - 1) (x^2 - 1) dx + -7 e integral_3^4 x dx + 11 e integral_3^4 1 dx

Expanding the integrand e^(x - 1) (x^2 - 1) gives e^(x - 1) x^2 - e^(x - 1):
= integral_3^4 (e^(x - 1) x^2 - e^(x - 1)) dx + -7 e integral_3^4 x dx + 11 e integral_3^4 1 dx

Integrate the sum term by term and factor out constants:
= integral_3^4 e^(x - 1) x^2 dx - integral_3^4 e^(x - 1) dx + -7 e integral_3^4 x dx + 11 e

integral_3^4 1 dx
For the integrand e^(x - 1) x^2, substitute u = x - 1 and du = dx.
This gives a new lower bound u = 3 - 1 = 2 and upper bound u = 4 - 1 = 3:
= integral_2^3 e^u (u + 1)^2 du - integral_3^4 e^(x - 1) dx + -7 e integral_3^4 x dx + 11 e

integral_3^4 1 dx
Expanding the integrand e^u (u + 1)^2 gives e^u u^2 + 2 e^u u + e^u:
= integral_2^3 (e^u u^2 + 2 e^u u + e^u) du - integral_3^4 e^(x - 1) dx + -7 e integral_3^4 x dx + 11 e integral_3^4 1 dx

Integrate the sum term by term and factor out constants:
= integral_2^3 e^u u^2 du + 2 integral_2^3 e^u u du + integral_2^3 e^u du - integral_3^4 e^(x - 1) dx + -7 e integral_3^4 x dx + 11 e integral_3^4 1 dx

For the integrand e^u u^2, integrate by parts, integral f dg = f g - integral g df, where
f = u^2, dg = e^u du, df = 2 u du, g = e^u:
= e^u u^2 right bracketing bar _2^3 + integral_2^3 e^u du - integral_3^4 e^(x - 1) dx + -7 e integral_3^4 x dx + 11 e integral_3^4 1 dx

Evaluate the antiderivative at the limits and subtract.
e^u u^2 right bracketing bar _2^3 = e^3 3^2 - e^2 2^2 = e^2 (9 e - 4):
= e^2 (9 e - 4) + integral_2^3 e^u du - integral_3^4 e^(x - 1) dx + -7 e integral_3^4 x dx + 11 e integral_3^4 1 dx

Apply the fundamental theorem of calculus.
The antiderivative of e^u is e^u:
= e^2 (9 e - 4) + e^u right bracketing bar _2^3 - integral_3^4 e^(x - 1) dx + -7 e integral_3^4 x dx + 11 e integral_3^4 1 dx

Evaluate the antiderivative at the limits and subtract.
e^u right bracketing bar _2^3 = e^3 - e^2 = (e - 1) e^2:
= (e - 1) e^2 + e^2 (9 e - 4) - integral_3^4 e^(x - 1) dx + -7 e integral_3^4 x dx + 11 e integral_3^4 1 dx

Apply the fundamental theorem of calculus.
The antiderivative of e^(x - 1) is e^(x - 1):
= (e - 1) e^2 + e^2 (9 e - 4) + (-e^(x - 1)) right bracketing bar _3^4 + -7 e integral_3^4 x dx + 11 e integral_3^4 1 dx

Evaluate the antiderivative at the limits and subtract.
(-e^(x - 1)) right bracketing bar _3^4 = (-e^(4 - 1)) - (-e^(3 - 1)) = -(e - 1) e^2:
= e^2 (9 e - 4) + -7 e integral_3^4 x dx + 11 e integral_3^4 1 dx

Apply the fundamental theorem of calculus.
The antiderivative of x is x^2/2:
= e^2 (9 e - 4) + (-(7 e x^2)/2) right bracketing bar _3^4 + 11 e integral_3^4 1 dx

Evaluate the antiderivative at the limits and subtract.
(-(7 e x^2)/2) right bracketing bar _3^4 = (-(7 e 4^2)/2) - (-(7 e 3^2)/2) = -(49 e)/2:
= -(49 e)/2 + e^2 (9 e - 4) + 11 e integral_3^4 1 dx

Apply the fundamental theorem of calculus.
The antiderivative of 1 is x:
= -(49 e)/2 + e^2 (9 e - 4) + 11 e x right bracketing bar _3^4

Evaluate the antiderivative at the limits and subtract.
11 e x right bracketing bar _3^4 = 11 e 4 - 11 e 3 = 11 e:
= e^2 (9 e - 4) - (27 e)/2
Which is equal to:
Answer: | = 1/2 e (-27 - 8 e + 18 e^2)

Guest Jun 2, 2017
#2
+202
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Thank you i find the same too but the result say its :

10e^3 - 5e^2 -49e/2 + 11e and i cant find it but is not false because it the result from ministry ... I dont know...Thank you after all!!!

Whitespy001  Jun 2, 2017
#3
+1

Just make sure that the Integral is bracketed accurately. Example: (e^x/e) - 7ex +11e dx? Or is it:

e^x/(e - 7ex + 11e) dx?

Guest Jun 2, 2017
#4
+202
+1

no its e^(x-1) and i write it e^x/e coz i cant write it there like this

Whitespy001  Jun 2, 2017
#5
+26542
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The result you quote isn't consistent with the integral you specify:

$$\int_3^4(x^2-1)e^{x-1}-7ex+11edx=9e^3-4e^2-\frac{27e}{2}$$

However:

$$\int_3^4x^2e^{x-1}-7ex+11edx=10e^3-5e^2-\frac{27e}{2}$$

Note that  -49e/2 + 11e  =  -27e/2

.

Alan  Jun 2, 2017
edited by Alan  Jun 2, 2017
#6
+202
+1

Yes thank you i know i find it but becaise the result is

10e^3 - 5e^2 -49e/2 + 11e from ministry i think i was false but im not .Thank you!

Whitespy001  Jun 2, 2017

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