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Math  /  Calculus

Question∫te4tdt\int t e^{4 t} d t

Studdy Solution

STEP 1

1. We are given the integral ∫te4t dt\int t e^{4t} \, dt.
2. We will use integration by parts to solve this integral.

STEP 2

1. Identify parts for integration by parts.
2. Differentiate and integrate the identified parts.
3. Apply the integration by parts formula.
4. Simplify the resulting expression.
5. Integrate any remaining terms.

STEP 3

Identify the parts for integration by parts. Let: - u=t u = t - dv=e4t dt dv = e^{4t} \, dt

STEP 4

Differentiate and integrate the identified parts: - Differentiate u u to get du=dt du = dt . - Integrate dv dv to get v=14e4t v = \frac{1}{4} e^{4t} .

STEP 5

Apply the integration by parts formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du:
∫te4t dt=t⋅14e4t−∫14e4t dt\int t e^{4t} \, dt = t \cdot \frac{1}{4} e^{4t} - \int \frac{1}{4} e^{4t} \, dt

STEP 6

Simplify the expression:
=t4e4t−14∫e4t dt= \frac{t}{4} e^{4t} - \frac{1}{4} \int e^{4t} \, dt

STEP 7

Integrate the remaining term ∫e4t dt\int e^{4t} \, dt:
∫e4t dt=14e4t+C\int e^{4t} \, dt = \frac{1}{4} e^{4t} + C
Substitute back:
=t4e4t−14(14e4t)+C= \frac{t}{4} e^{4t} - \frac{1}{4} \left( \frac{1}{4} e^{4t} \right) + C

STEP 8

Simplify the final expression:
=t4e4t−116e4t+C= \frac{t}{4} e^{4t} - \frac{1}{16} e^{4t} + C
Combine terms:
=(t4−116)e4t+C= \left( \frac{t}{4} - \frac{1}{16} \right) e^{4t} + C
The solution to the integral is:
(t4−116)e4t+C\boxed{\left( \frac{t}{4} - \frac{1}{16} \right) e^{4t} + C}

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