Advertisements
Advertisements
Question
Advertisements
Solution
\[\text{We have}, \]
\[I = \int_0^2 \left[ x \right] d x\]
\[\text{We know that}, \]
\[\left[ x \right] = \begin{cases}0&,& 0 < x < 1\\1&,& 1 < x < 2\end{cases}\]
\[ \therefore I = \int_0^2 \left[ x \right] d x\]
\[ = \int_0^1 \left[ x \right] d x + \int_1^2 \left[ x \right] d x\]
\[ = \int_0^1 \left( 0 \right) d x + \int_1^2 \left( 1 \right) d x\]
\[ = 0 + \left[ x \right]_1^2 \]
\[ = 2 - 1 = 1\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following definite integrals:
Evaluate :
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
Which notation represents a definite integral?
