Advertisements
Advertisements
प्रश्न
If `int_1^"a" (3x^2 + 2x + 1) "d"x` = 11, find the real value of a
उत्तर
Given, `int_1^"a" (3x^2 + 2x + 1) "d"x` = 11
∴ `[(3x^3)/3 + (2x^2)/2 + x]_1^"a"` = 11
∴ `[x^3 + x^2 + x]_1^"a"` = 11
∴ (a3 + a2 + a) – (1 + 1 + 1) = 11
∴ a3 + a2 + a – 3 = 11
∴ a3 + a2 + a – 14 = 0
∴ (a – 2)(a2 + 3a + 7) = 0
∴ a = 2 or a2 + 3a + 7 = 0
But, a2 + 3a + 7 = 0 does not have real roots.
∴ a = 2
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_(-4)^2 (1)/(x^2 + 4x + 13)*dx`
Evaluate:
`int_0^1 (1)/sqrt(3 + 2x - x^2)*dx`
Evaluate : `int_0^(pi//4) (sin2x)/(sin^4x + cos^4x)*dx`
Evaluate the following : `int_(-3)^(3) x^3/(9 - x^2)*dx`
`int_0^(log5) (e^x sqrt(e^x - 1))/(e^x + 3) * dx` = ______.
Evaluate the following definite integrals: if `int_1^"a" (3x^2 + 2x + 1)*dx` = 11, find a.
Evaluate the following integrals : `int_(-9)^9 x^3/(4 - x^2).dx`
Evaluate the following integrals : `int_1^2 sqrt(x)/(sqrt(3 - x) + sqrt(x))*dx`
Choose the correct alternative :
`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))*dx` =
Solve the following : `int_0^1 (x^2 + 3x + 2)/sqrt(x)*dx`
`int_0^1 sqrt((1 - x)/(1 + x)) "d"x` =
Choose the correct alternative:
`int_2^3 x^4 "d"x` =
Evaluate `int_0^1 "e"^(x^2)*"x"^3 "d"x`
`int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2))` dx = ?
`int_2^3 "x"/("x"^2 - 1)` dx = ____________.
Evaluate the following definite intergral:
`int_1^3 logx dx`
`int_a^b f(x) dx = int_a^b f (t) dt`
Evaluate the following definite integral:
`int_4^9 1/sqrtx dx`
Evaluate the following definite intergral:
`int_4^9 1/sqrtxdx`
Evaluate the following definite intergral:
`int_1^2(3x)/((9x^2-1))dx`
