Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Advertisements
उत्तर
`int_1^4` f(x) dx = `int_1^2 "f"(x) "d"x + int_2^4 "f"(x) "d"x`
= `int_1^2 (4x + 3) "d"x + int_2^4 (3x + 5) "d"x`
= `[4(x^2/2) + 3x]_1^2 + [3(x^2/2) + 5x]_2^4`
= `[2x^2 +3x]_1^2 + [3/2 x^2 + 5x]_2^4`
= `[2(2)^2 + 3(2)] - [2(1)^2 + 3(1)] + [3/2(4)^2 + 5(4)] - [3/2 (2)^2+ 5(2)]`
= `[2(4) + 6] - [2 + 3] + [3/ (16) + 20] - [3/2 (4) + 10]`
= [8 + 6] – [5] + [24 + 20] – [6 + 10]
= 14 – 5 + 44 – 16
= 58 – 21
= 37
APPEARS IN
संबंधित प्रश्न
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
