If _-0.15^0.15|100x^2-1|,dx = k3000, then k is equal to ______.

Question

If $\displaystyle\int_{-0.15}^{0.15}|100x^2-1|\,dx = \dfrac{k}{3000}$, then $k$ is equal to \_\_\_\_\_\_.

Answer: 575

Accepted Answer

By symmetry, $I=2\displaystyle\int_0^{0.15}|100x^2-1|\,dx$. Since $100x^2-1=0$ at $x=0.1$:

$$I = 2\int_0^{0.1}(1-100x^2)\,dx + 2\int_{0.1}^{0.15}(100x^2-1)\,dx$$

$$= 2\left[x-\frac{100x^3}3\right]_0^{0.1} + 2\left[\frac{100x^3}3-x\right]_{0.1}^{0.15}$$

$$= 2\left(0.1-\frac{0.1}3\right) + 2\left(0.1125-0.15-\frac{0.1}3+0.1\right) = \frac{575}{3000}$$

So $k=575$.

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