Answer :
We start with the integral
[tex]$$
\int_0^1 7x^6 e^{x^7}\,dx.
$$[/tex]
A useful technique here is substitution. Let
[tex]$$
u = x^7.
$$[/tex]
Then, the derivative of [tex]$u$[/tex] with respect to [tex]$x$[/tex] is
[tex]$$
\frac{du}{dx} = 7x^6 \quad \Longrightarrow \quad du = 7x^6\,dx.
$$[/tex]
Notice that the differential [tex]$7x^6\,dx$[/tex] in the original integral is exactly [tex]$du$[/tex]. Also, we must change the limits of integration. When [tex]$x=0$[/tex],
[tex]$$
u = 0^7 = 0,
$$[/tex]
and when [tex]$x=1$[/tex],
[tex]$$
u = 1^7 = 1.
$$[/tex]
Thus, after substitution, the integral becomes
[tex]$$
\int_0^1 e^u\,du.
$$[/tex]
The antiderivative of [tex]$e^u$[/tex] is [tex]$e^u$[/tex]. Evaluating the integral from [tex]$u=0$[/tex] to [tex]$u=1$[/tex], we have
[tex]$$
e^u \Big|_0^1 = e^1 - e^0 = e - 1.
$$[/tex]
Therefore, the value of the original integral is
[tex]$$
e - 1,
$$[/tex]
which numerically is approximately [tex]$1.718281828459045$[/tex].
[tex]$$
\int_0^1 7x^6 e^{x^7}\,dx.
$$[/tex]
A useful technique here is substitution. Let
[tex]$$
u = x^7.
$$[/tex]
Then, the derivative of [tex]$u$[/tex] with respect to [tex]$x$[/tex] is
[tex]$$
\frac{du}{dx} = 7x^6 \quad \Longrightarrow \quad du = 7x^6\,dx.
$$[/tex]
Notice that the differential [tex]$7x^6\,dx$[/tex] in the original integral is exactly [tex]$du$[/tex]. Also, we must change the limits of integration. When [tex]$x=0$[/tex],
[tex]$$
u = 0^7 = 0,
$$[/tex]
and when [tex]$x=1$[/tex],
[tex]$$
u = 1^7 = 1.
$$[/tex]
Thus, after substitution, the integral becomes
[tex]$$
\int_0^1 e^u\,du.
$$[/tex]
The antiderivative of [tex]$e^u$[/tex] is [tex]$e^u$[/tex]. Evaluating the integral from [tex]$u=0$[/tex] to [tex]$u=1$[/tex], we have
[tex]$$
e^u \Big|_0^1 = e^1 - e^0 = e - 1.
$$[/tex]
Therefore, the value of the original integral is
[tex]$$
e - 1,
$$[/tex]
which numerically is approximately [tex]$1.718281828459045$[/tex].
