find value of def integral with a=-2 and b=2sqrt(3) definite integral is : x^3 * sqrt(x^2+4) dx
for integral i get 1/15 *((4+x^2)^(3/2)) (-8+3x^2)
for value i get [1536- 64sqrt(2)]/15 but its' wrong. help please
To find the value of the definite integral with the given values of a and b, you first need to substitute the values of a and b into the antiderivative expression and then subtract the result of substituting a from the result of substituting b.
Let's start by finding the antiderivative of the integrand expression:
∫(x³*√(x²+4)) dx
To solve this integral, you can perform a substitution. Let u = x² + 4, then du = 2x dx. Rearranging gives dx = du / (2x).
Substituting these values into the original integral:
∫(x³*√(x²+4)) dx = ∫((x²*2x/2x) * √(x²+4)) dx
= ∫((x²/2) * 2√(x²+4) * (1/x)) dx
= ∫(x * √(x²+4)) dx
= ∫(√u) du
= (2/3) * u^(3/2) + C
= (2/3) * (x²+4)^(3/2) + C
Now, substitute the values of a and b into this antiderivative expression and subtract the result of substituting a from the result of substituting b:
F(b) - F(a) = [(2/3) * (b²+4)^(3/2)] - [(2/3) * (a²+4)^(3/2)]
Substituting a = -2 and b = 2√3:
F(2√3) - F(-2) = [(2/3) * ((2√3)²+4)^(3/2)] - [(2/3) * ((-2)²+4)^(3/2)]
= [(2/3) * (12+4)^(3/2)] - [(2/3) * (4+4)^(3/2)]
= [(2/3) * (16)^(3/2)] - [(2/3) * (8)^(3/2)]
= (2/3) * 64 - (2/3) * 16
= (128/3) - (32/3)
= 96/3
= 32
Therefore, the correct value of the definite integral is 32.