Advertisements
Advertisements
प्रश्न
Solve the following : `int_1^2 x^2*dx`
उत्तर
Let I = `int_1^2 x^2*dx`
= `[x^3/3]_1^2`
= `(1)/(3) (2^3 - 1^3)`
= `(1)/(3)(8 - 1)`
∴ I = `(7)/(3)`.
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_(-4)^2 (1)/(x^2 + 4x + 13)*dx`
Evaluate:
`int_0^(pi/2) sqrt(cos x) sin^3x * dx`
Evaluate : `int_0^(pi/2) (1)/(5 + 4 cos x)*dx`
Evaluate the following : `int_((-pi)/4)^(pi/4) x^3 sin^4x*dx`
Evaluate the following : `int_0^pi x sin x cos^2x*dx`
Evaluate the following : `int_(-1)^(1) (1 + x^3)/(9 - x^2)*dx`
Evaluate the following integrals : `int_1^2 sqrt(x)/(sqrt(3 - x) + sqrt(x))*dx`
State whether the following is True or False : `int_"a"^"b" f(x)*dx = int_"a"^"b" f("t")*dt`
Solve the following : `int_2^3 x/(x^2 - 1)*dx`
Solve the following : `int_0^4 (1)/sqrt(x^2 + 2x + 3)*dx`
Prove that: `int_"a"^"b" "f"(x) "d"x = int_"a"^"c""f"(x) "d"x + int_"c"^"b" "f"(x) "d"x`, where a < c < b
Evaluate:
`int_1^2 1/(x^2 + 6x + 5) dx`
Evaluate `int_1^2 "e"^(2x) (1/x - 1/(2x^2)) "d"x`
`int_2^3 "x"/("x"^2 - 1)` dx = ____________.
Evaluate the following definite intergral:
`int_1^3 logx dx`
Evaluate the following definite integral:
`int_1^3 log x dx`
Evaluate the following integrals:
`int_-9^9 (x^3)/(4 - x^2) dx`
Evaluate the following definite intergral:
`int _1^3logxdx`
Evaluate the following definite integral:
`int_1^3 logx dx`
Evaluate the following definite intergral:
`int_-2^3 1/(x+5)dx`
