CBSE BoardEnglish MediumSTD 12 ScienceApplied MathsIntegrals and it's Applications2 Marks
Question
Evaluate: $\int_1^2 \frac{x^3-1}{x^2} d x$
✓
Answer
Let $I=\int_1^2\left(\frac{x^3}{x^2}-\frac{1}{x^2}\right) d x$ $\begin{aligned} & =\int_1^2\left(x-\frac{1}{x^2}\right) d x \\ & =\left[\frac{x^2}{2}+\frac{1}{x}\right]_1^2 \\ I & =\left(2+\frac{1}{2}\right)-\left(\frac{1}{2}+1\right) \\ & =\frac{5}{2}-\frac{3}{2} \\ I & =1\end{aligned}$
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