Advertisements
Advertisements
Question
Advertisements
Solution
\[\text{We have}, \]
\[I = \int_0^1 e^\left\{ x \right\} d x\]
\[\text{We know that}, \]
\[\left\{ x \right\} = x\text{, when }0 < x < 1\]
\[ \therefore I = \int_0^1 e^x d x\]
\[ = \left[ e^x \right]_0^1 \]
\[ = e^1 - e^0 \]
\[ = e - 1\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate the following integral:
Evaluate each of the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
