Find the value of $\int^5_0f(x)dx$

Question

Let $f(-x)=2f(x)$. If $\int^5_{-5}f(x)dx=3$, then $\int^5_0f(x)dx$ is...

(A) -5

(B) -3

(C) -1

(D) 0

(E) 1

My attempt:

$\begin{aligned} f(x)&=\frac{f(-x)}{2}\\ \int^5_{-5}f(x)dx&=\int^0_{-5}f(x)dx+\int^5_0f(x)dx\\ &=\frac{-1}{2}\int^0_{-5}f(-x)dx+\int^5_0f(x)dx\\ \text{Let }u=-x \text{, then }du=-dx\\ \int^5_{-5}f(x)dx&=\frac{1}{2}\int^5_{0}f(u)du+\int^5_0f(x)dx\\ &=\frac{1}{2}\int^5_{0}f(x)dx+\int^5_0f(x)dx\\ &=\frac{3}{2}\int^5_0f(x)dx\\\\ \frac{3}{2}\int^5_0f(x)dx&=3\\ \int^5_0f(x)dx&=2 \end{aligned}$

The answer is not on the option.

Answer 1

Split up the integral
$$ \int_{-5}^5 f(x)dx = \int_{-5}^0 f(x)dx + \int_{0}^5 f(x)dx = 3 $$

We change the bounds of the negative integral and apply the hypothesis

$$ \int_{-5}^0 f(x)dx = \int_{0}^5 f(-x)dx = \int_{0}^5 2 f(x)dx = 2 \int_{0}^5 f(x)dx $$

Ergo $$ 3 = \int_{-5}^0 f(x)dx + \int_{0}^5 f(x)dx = 2\int_{0}^5 f(x)dx + \int_{0}^5 f(x)dx = 3 \int_{0}^5 f(x)dx $$

We conclude that $\int_{0}^5 f(x)dx = 1$

Knowing the answer, can you find the mistake in your original reasoning?