Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
बेरीज
Advertisements
उत्तर
= `int_0^(1/4) sqrt((1 - 4)^(1/2)) "d"x`
= `[(1 - 4x)^(3/2)/((3/2)(-4))]_0^(1/4)`
= `[(1 - 4x)^(3/2)/(-6)]_0^(1/4)`
= `- 1/6 [(1 - 4x)^(3/2)]_0^(1/4)`
= `- 1/6 [(1 - 4(1/4))^(3/2) - [1 - 4(0)]^(3/2)]`
= `- 1/6 [0 - (1)^(3/2)]`
= `- 1/6 (- 1)`
= `1/6`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/2} x^2 \cos\ 2x\ dx\]
\[\int\limits_1^2 \frac{1}{x \left( 1 + \log x \right)^2} dx\]
\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
\[\int\limits_2^3 \left( 2 x^2 + 1 \right) dx\]
\[\int\limits_2^3 \frac{1}{x}dx\]
Solve each of the following integral:
\[\int_2^4 \frac{x}{x^2 + 1}dx\]
\[\int\limits_0^1 \frac{x}{\left( 1 - x \right)^\frac{5}{4}} dx =\]
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
