Evaluate the integral
..x
∫.....(t+7)^2 dt
..0
(1/3) * ( (x + 7)^3 - 7^3 )
0x? as in zero times x which would be zero and then that canceling out the other part of the integrand? o.O I'm confusioned as to what you are integrating xD... but let me take the integral of (t+7)^2 dt
â«(t+7)^2 dt Simple substitution
u = t + 7
du = dt
â«(u)^2 du
(u^3)/3
((t + 7)^3)/3 + c
Hope that helped :\
I = ⫠( t ² + 14 t + 49 dt ---------limits 0 to x
I = [ t ³ / 3 + 7 t ² + 49 t ] ---------limits 0 to x
I = x ³ / 3 + 7 x ² + 49 x
f'(t) = (t + 7)^2
f(t) = (1/3) * (t + 7)^3 + C
f(x) - f(0) =
(1/3) * (x + 7)^3 + C - (1/3) * (0 + 7)^3 - C
(1/3) * ( (x + 7)^3 - 7^3 ) =
(1/3) * (x + 7 - 7) * ((x + 7)^2 + 2 * 7 * (x + 7) + 7^2)
(1/3) * (x) * (x^2 + 14x + 49 + 14x + 98 + 49)
(1/3) * x * (x^2 + 28x + 196)
(1/3) * (x^3 + 28x^2 + 196x)
hint: expand the integrand.
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(1/3) * ( (x + 7)^3 - 7^3 )
0x? as in zero times x which would be zero and then that canceling out the other part of the integrand? o.O I'm confusioned as to what you are integrating xD... but let me take the integral of (t+7)^2 dt
â«(t+7)^2 dt Simple substitution
u = t + 7
du = dt
â«(u)^2 du
(u^3)/3
((t + 7)^3)/3 + c
Hope that helped :\
I = ⫠( t ² + 14 t + 49 dt ---------limits 0 to x
I = [ t ³ / 3 + 7 t ² + 49 t ] ---------limits 0 to x
I = x ³ / 3 + 7 x ² + 49 x
f'(t) = (t + 7)^2
f(t) = (1/3) * (t + 7)^3 + C
f(x) - f(0) =
(1/3) * (x + 7)^3 + C - (1/3) * (0 + 7)^3 - C
(1/3) * ( (x + 7)^3 - 7^3 ) =
(1/3) * (x + 7 - 7) * ((x + 7)^2 + 2 * 7 * (x + 7) + 7^2)
(1/3) * (x) * (x^2 + 14x + 49 + 14x + 98 + 49)
(1/3) * x * (x^2 + 28x + 196)
(1/3) * (x^3 + 28x^2 + 196x)
hint: expand the integrand.