Advertisements
Advertisements
Question
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
Options
a = `(-1)/8`, b = `7/8`
a = `1/8`, b = `7/8`
a = `(-1)/8`, b = `(-7)/8`
a = `1/8`, b = `(-7)/8`
Solution
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then a = `(-1)/8`, b = `(-7)/8`.
Explanation:
`(3"e"^x - 5"e"^-x)/(4"e"^x + 5"e"^-x) = "a" + "b" ((4"e"^x - 5"e"^-x))/(4"e"^x + 5"e"^-x)`,
Giving 3ex – 5e –x = a(4ex + 5e–x) + b(4ex – 5e–x).
Comparing coefficients on both sides,
We get 3 = 4a + 4b and –5 = 5a – 5b.
This verifies a = `(-1)/8`, b = `7/8`.
APPEARS IN
RELATED QUESTIONS
Evaluate the following definite integrals:
If f(x) is a continuous function defined on [−a, a], then prove that
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
