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Question:
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$$
\int_0^{2 a} f(x) d x=
$$
Options:
\int_0^{2 a} f(x) d x=
$$
Solution:
1396 Upvotes
Verified Answer
The correct answer is:
$\int_0^a(f(x)+f(x+a)) d x$
Given $\int_0^{2 a} f(x) d x$
$$
\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_a^{2 a} f(x) d x
$$
Let $x=a+t$, when $x=a, t=0$ $d x=d t, x=2 a, t=a$
$$
\begin{aligned}
& \int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_a^{2 a} f(a+t) d t \\
& \int_0^a f(x) d x+\int_0^a f(a+x) d x \\
& =\int_0^a[f(x)+f(a+x)] d x
\end{aligned}
$$
So, option (b) is correct.
$$
\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_a^{2 a} f(x) d x
$$
Let $x=a+t$, when $x=a, t=0$ $d x=d t, x=2 a, t=a$
$$
\begin{aligned}
& \int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_a^{2 a} f(a+t) d t \\
& \int_0^a f(x) d x+\int_0^a f(a+x) d x \\
& =\int_0^a[f(x)+f(a+x)] d x
\end{aligned}
$$
So, option (b) is correct.
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