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NDA 2013
Exam
NDA
Year
2013
Subject
Mathematics
Chapter
Definite Integration

NDA 2013 · Mathematics · Definite Integration

What is ∫_1^2 x d x equal to?

What is ∫12ln⁡xdx\int_1^2 \ln x d x equal to?

Options

  1. A\ln 2
  2. B1
  3. Cln⁡(4e)\ln \left(\frac{4}{e}\right)
  4. Dln⁡(e4)\ln \left(\frac{e}{4}\right)

Correct answer

C. ln⁡(4e)\ln \left(\frac{4}{e}\right)

Step-by-step solution

I=∫12ln⁡(x)⋅dx\mathrm{I}=\int_1^2 \ln (x) \cdot d x From integration by parts formula u=ln⁡(x),v=1∫uvdx=u∫v⋅dx−∫(dudx⋅∫v⋅dx)dx=ln⁡x∫12dx−∫12(ddx(ln⁡x)⋅∫dx)⋅dx=[ln⁡x⋅(x)]12−∫121x⋅x⋅dx=[x⋅ln⁡(x)−x]12=2⋅ln⁡(2)−2−1⋅ln⁡1+1=ln⁡(2)2−1=ln⁡(4)−ln⁡e=ln⁡(4e)\begin{aligned} & u=\ln (x), \quad v=1 \\ & \int u v d x=u \int v \cdot d x-\int\left(\frac{d u}{d x} \cdot \int v \cdot d x\right) d x \\ = & \ln x \int_1^2 d x-\int_1^2\left(\frac{d}{d x}(\ln x) \cdot \int d x\right) \cdot d x \\ = & {[\ln x \cdot(x)]_1^2-\int_1^2 \frac{1}{x} \cdot x \cdot d x } \\ = & {[x \cdot \ln (x)-x]_1^2 } \\ = & 2 \cdot \ln (2)-2-1 \cdot \ln 1+1 \\ = & \ln (2)^2-1 \\ = & \ln (4)-\ln e=\ln \left(\frac{4}{e}\right) \end{aligned}
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