Advertisements
Advertisements
Question
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
Advertisements
Solution
\[Let I = \int_1^5 \frac{x}{\sqrt{2x - 1}} d x\]
\[Let, 2x - 1 = t,\text{ then }2dx = dt, \]
\[\text{When, }x \to 1 ; t \to 1\text{ and x to 5; } t \to 9\]
\[x = \frac{t + 1}{2}\]
\[I = \frac{1}{2} \int_1^9 \frac{t + 1}{\sqrt{t}} \times \frac{dt}{2}\]
\[ = \frac{1}{4} \left[ \frac{2 t^\frac{3}{2}}{3} + 2\sqrt{t} \right]_1^9 \]
\[ = \frac{1}{4}\left[ 18 + 6 - \frac{2}{3} - 2 \right]\]
\[ = \frac{16}{3}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
What is the result of a definite integral?
