Advertisements
Advertisements
Question
Options
1
e − 1
e + 1
0
Advertisements
Solution
1
\[\int_1^e \log x d x\]
\[ = \int_1^e \log x x^0 d x\]
\[ = \left[ x \log x \right]_1^e - \int_1^e \frac{1}{x}x d x\]
\[ = \left[ x \log x \right]_1^e - \left[ x \right]_1^e \]
\[ = \left( e - 0 \right) - \left( e - 1 \right)\]
\[ = e - e + 1\]
\[ = 1\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
