evaluate the iterated integral. 1 0 s6 cos(s7) dt ds 0

Answers

Answer 1

The value of the evaluated iterated integral is 0.1202.

As per the given data the iterated integral is [tex]I & =\int_0^1 \int_0^{s^6} \cos \left(s^7\right) d t d s[/tex].

Here we have to evaluate the given iterated integral.

The definite integral of a real-valued function f(x) with respect to a real variable x on an interval [a, b] is expressed as[tex]$\int_a^b f(x) d x$[/tex] = F(b) - F(a)

Here,

[tex]$\int=$[/tex] Integration symbol

a = Lower limit

b = Upper limit

f(x) = Integrand

dx = Integrating agent

Thus, [tex]$\int a b f(x) d x$[/tex] is read as the definite integral of f(x) with respect to d x from a to b.

Indefinite integrals are implemented when the boundaries of the integrand are not specified.

[tex]I & =\int_0^1 \int_0^{s^6} \cos \left(s^7\right) d t d s[/tex]

Integrate the given iterated integral.

We get:

[tex]& =\int_0^1 \cos \left(s^7\right)\left(\int_0^{s^6} d t\right) d s[/tex]

[tex]& =\int_0^1 \cos \left(s^7\right) \times[t]_0^{s^6} d s . \\[/tex]

[tex]& =\int_0^1 \cos \left(s^7\right) \times\left(s^6-0\right) d s . \\[/tex]

[tex]& =\int_0^1 s^6 \cos \left(s^7\right) d s[/tex]

Say [tex]s^7=t[/tex]                 when s = 0

[tex]\Rightarrow \frac{d t}{d s}=7 s^6 \quad$[/tex]                          t = 0  

[tex]\Rightarrow \frac{d t}{7}=s^6 d s[/tex]              when s = 1.....   t = 1

Replace the assumed terms.

Then we get:

[tex]$I=\int_0^1 \cos (t) \frac{d t}{7}$[/tex]

[tex]$=\frac{1}{7} \int_0^1 \cos (t) d t$[/tex]

[tex]$=\frac{1}{7}[\sin (t)]_0^1=\frac{1}{7}[\sin (1)-\sin (0)]$[/tex]

[tex]=\frac{\sin (1)}{7}$[/tex]

[tex]\int_0^1 \int_0^{s^6} \cos \left(s^7\right) d t d s[/tex] = 0.1202

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