$$ \int _ { 1 } ^ { 4 } \int _ { 0 } ^ { 1 u } \cos ( u y ) d y d u $$

Question

UpStudy AI · Accepted Answer

We begin with the integral

$$
\int_{1}^{4} \int_{0}^{\frac{1}{u}} \cos(u y) \, dy \, du.
$$

### Step 1. Evaluate the inner integral

Consider the inner integral

$$
I(u) = \int_{0}^{\frac{1}{u}} \cos(u y) \, dy.
$$

Make the substitution

$$
v = u y, \quad \text{so that} \quad dv = u \, dy \quad \Rightarrow \quad dy = \frac{dv}{u}.
$$

When $$ y = 0 $$, then $$ v = 0 $$; and when $$ y = \frac{1}{u} $$, then $$ v = 1 $$. The integral becomes

$$
I(u) = \int_{0}^{1} \cos(v) \cdot \frac{dv}{u} = \frac{1}{u} \int_{0}^{1} \cos(v) \, dv.
$$

Now, compute the integral

$$
\int_{0}^{1} \cos(v) \, dv = \sin(v) \Big|_{0}^{1} = \sin(1) - \sin(0) = \sin(1).
$$

Thus,

$$
I(u) = \frac{\sin(1)}{u}.
$$

### Step 2. Evaluate the outer integral

Substitute $$ I(u) $$ in the outer integral:

$$
\int_{1}^{4} \frac{\sin(1)}{u} \, du = \sin(1) \int_{1}^{4} \frac{1}{u} \, du.
$$

We know that

$$
\int \frac{1}{u} \, du = \ln |u| + C.
$$

So,

$$
\int_{1}^{4} \frac{1}{u} \, du = \ln|u|\Big|_{1}^{4} = \ln(4) - \ln(1) = \ln(4),
$$

since $$ \ln(1) = 0 $$.

### Final Answer

Multiplying by $$ \sin(1) $$, the value of the original integral is

$$
\sin(1) \ln(4).
$$
The value of the integral is $$ \sin(1) \ln(4) $$.

\( \int _ { 1 } ^ { 4 } \int _ { 0 } ^ { 1 u } \cos ( u y ) d y d u \)