Evaluate. (Be sure to check by differentiating!)
$\int\left(6 t^4+8\right) t^3 d t$
C. $u=6 t+8$
D. $u=6 t^4+8$
Write the integral in terms of $u$.
$\int\left(6 t^4+8\right) t^3 d t=\int\left(\frac{u}{24}\right) d u$
(Type an exact answer. Use parentheses to clearly denote the argument of each function.)
Evaluate the integral.
$\int\left(6 t^4+8\right) t^3 d t=$
(Type an exact answer. Use parentheses to clearly denote the argument of each function.)
Answer (1)
Find d u / d t using the substitution u = 6 t 4 + 8 , which gives d u / d t = 24 t 3 .
Substitute u and d t = d u / ( 24 t 3 ) into the integral, transforming it to ∫ ( u /24 ) d u .
Evaluate the integral with respect to u : 24 1 ∫ u d u = 48 u 2 + C .
Substitute back u = 6 t 4 + 8 and simplify to get the final answer: 4 3 t 8 + 2 t 4 + C .
Explanation
Problem Analysis We are given the integral ∫ ( 6 t 4 + 8 ) t 3 d t and the substitution u = 6 t 4 + 8 . We want to evaluate the integral and express the result in terms of t .
Finding du and Substituting First, we find the derivative of u with respect to t :
d t d u = d t d ( 6 t 4 + 8 ) = 24 t 3 From this, we can express d t in terms of d u :
d t = 24 t 3 d u Now we substitute u and d t into the integral: ∫ ( 6 t 4 + 8 ) t 3 d t = ∫ u t 3 24 t 3 d u = ∫ 24 u d u
Integrating with respect to u Next, we evaluate the integral with respect to u :
∫ 24 u d u = 24 1 ∫ u d u = 24 1 ⋅ 2 u 2 + C = 48 u 2 + C
Substituting back and Simplifying Now we substitute back u = 6 t 4 + 8 to express the result in terms of t :
48 u 2 + C = 48 ( 6 t 4 + 8 ) 2 + C = 48 ( 6 t 4 + 8 ) ( 6 t 4 + 8 ) + C = 48 36 t 8 + 96 t 4 + 64 + C = 48 36 t 8 + 48 96 t 4 + 48 64 + C = 4 3 t 8 + 2 t 4 + 3 4 + C Since 3 4 is a constant, we can absorb it into the constant of integration, so we can write the result as: 4 3 t 8 + 2 t 4 + C
Final Result Therefore, the integral evaluates to: ∫ ( 6 t 4 + 8 ) t 3 d t = 4 3 t 8 + 2 t 4 + C
Examples
Imagine you're calculating the total energy consumption of a device where the power consumption varies with time according to the function ( 6 t 4 + 8 ) t 3 . Evaluating the integral of this function over a period of time gives you the total energy consumed. This is a practical application of integration in engineering and physics, helping to understand and optimize energy usage in various systems.
