Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
योग
Advertisements
उत्तर
`int_0^oo "e"^(-mx) x^6 "d"x = int_0^oo x^"n""e"^(-"a"x) "d"x`
= `("n"!)/("a"^("n" + 1)`
Where n = 6
a = m
So the integral becomes = `(6!)/(3^(6 + 1)) = (6!)/"m"^7`
shaalaa.com
Definite Integrals
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Integral Calculus – 1 - Exercise 2.10 [पृष्ठ ५१]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/2} \left( \sin x + \cos x \right) dx\]
\[\int\limits_{- 1}^1 \log\left( \frac{2 - x}{2 + x} \right) dx\]
\[\int\limits_0^1 \log\left( \frac{1}{x} - 1 \right) dx\]
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
\[\int\limits_0^5 \left( x + 1 \right) dx\]
Evaluate each of the following integral:
\[\int_e^{e^2} \frac{1}{x\log x}dx\]
\[\int\limits_0^\pi \frac{1}{a + b \cos x} dx =\]
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
