Calculate the iterated integral: $\int_0^1 \int_0^1 27(u-v)^7 \, du \, dv$

Neetesh Kumar | November 29, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 15.1 Question Number 10
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Step-by-step solution:

Step 1: Evaluate the Inner Integral

The inner integral is:

$\int_0^1 27(u-v)^7 \, du$

Since $v$ is treated as a constant with respect to $u$, we can factor out the constant $27$:

$27 \int_0^1 (u-v)^7 \, du$

Now, let's solve the integral of $(u - v)^7$. The antiderivative of $(u - v)^7$ with respect to $u$ is:

$\frac{(u - v)^8}{8}$

So, we evaluate the definite integral from $u = 0$ to $u = 1$:

$27 \left[ \frac{(u - v)^8}{8} \right]_0^1$

Evaluating at the bounds:

• At $u = 1$: $\frac{(1 - v)^8}{8}$
• At $u = 0$: $\frac{(0 - v)^8}{8} = \frac{-v^8}{8} = \frac{v^8}{8}$

Thus, the result of the inner integral is:

$27 \left( \frac{(1 - v)^8}{8} - \frac{v^8}{8} \right)$

Factor out the $\frac{27}{8}$:

$\frac{27}{8} \left( (1 - v)^8 - v^8 \right)$

Step 2: Evaluate the Outer Integral

Now, we evaluate the outer integral:

$\int_0^1 \frac{27}{8} \left( (1 - v)^8 - v^8 \right) dv$

We can break this into two separate integrals:

$\frac{27}{8} \left( \int_0^1 (1 - v)^8 \, dv - \int_0^1 v^8 \, dv \right)$

First Integral: Evaluate $\int_0^1 (1 - v)^8 \, dv$

Let $w = 1 - v$, so $dw = -dv$. The limits change accordingly: when $v = 0$, $w = 1$, and when $v = 1$, $w = 0$. Thus, the integral becomes:

$\int_0^1 (1 - v)^8 \, dv = \int_1^0 w^8 (-dw) = \int_0^1 w^8 \, dw$

The antiderivative of $w^8$ is:

$\frac{w^9}{9}$

Evaluating from $w = 0$ to $w = 1$:

$\frac{w^9}{9} \Big|_0^1 = \frac{1^9}{9} - \frac{0^9}{9} = \frac{1}{9}$

Second Integral: Evaluate $\int_0^1 v^8 \, dv$

The antiderivative of $v^8$ is:

$\frac{v^9}{9}$

Evaluating from $v = 0$ to $v = 1$:

$\frac{v^9}{9} \Big|_0^1 = \frac{1^9}{9} - \frac{0^9}{9} = \frac{1}{9}$

Step 3: Combine the Results

Now, substitute the results of the two integrals back into the outer integral:

$\frac{27}{8} \left( \frac{1}{9} - \frac{1}{9} \right)$

This simplifies to:

$\frac{27}{8} \times 0 = 0$

Final Answer:

The value of the iterated integral is: $\boxed{0}$

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