Practicing Success
$\int_1^3|(2-x)\log_ex|dx$ is equal to: |
$\frac{3}{2}\log_e3+\frac{1}{2}$ $\log_e\frac{16}{3\sqrt{2}}-\frac{1}{2}$ $-\frac{3}{2}\log_e3-\frac{1}{2}$ None of these |
None of these |
$\int\limits_1^3|(2-x)\log_ex|dx=\int\limits_1^3|(2-x)|\log_ex\,dx$ [∴ loge x > 0 ∀ x ∈ (1,3)] = $\int\limits_1^2(2-x)\log x\,dx+\int\limits_2^3(x-2)\log x\,dx$ Use : $\int\log x\,dx=x\,\log x-x+c$ and $\int x\,\log x\,dx=\frac{x^2}{2}\log x-\frac{x^2}{4}+c$ $⇒I=2[x\,\log x-x]_1^2-[\frac{x^2}{2}\log x]_1^2+[\frac{x^2}{4}]_1^2+[x^2\frac{\log x}{2}]_2^3-[\frac{x^2}{4}]_2^3-[2x\,\log x]_2^3+2[x]_2^3$ $=4\,ln\,2-\frac{3}{2}ln\,3-\frac{1}{2}=ln\frac{16}{3}\sqrt{3}-\frac{1}{2}$ |