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प्रश्न
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
पर्याय
1
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MCQ
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उत्तर
2
shaalaa.com
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संबंधित प्रश्न
\[\int\limits_0^1 \frac{x}{x + 1} dx\]
\[\int\limits_0^{\pi/2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) dx\]
\[\int\limits_0^{\pi/2} x \cos\ x\ dx\]
\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot x} dx\]
\[\int\limits_0^a \frac{1}{x + \sqrt{a^2 - x^2}} dx\]
If f(x) is a continuous function defined on [−a, a], then prove that
\[\int\limits_{- a}^a f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( - x \right) \right\} dx\]
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
