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Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 67

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Answer:  \frac{62}{3}

Hint: To solve the given statement we have to integrate them individually.

Given: \int_{1}^{3}\left(x^{2}+3 x\right) d x


\begin{aligned} &\int_{1}^{3}\left(x^{2}+3 x\right) d x \\ &\int_{1}^{3}\left(x^{2}+3 x\right) d x \\ &=\int_{1}^{3} x^{2} d x+\int_{1}^{3} 3 x d x \end{aligned}

\begin{aligned} &=\left[\frac{x^{3}}{3}\right]_{1}^{3}+3\left[\frac{x^{2}}{2}\right]_{1}^{3} \\ &=\frac{27-1}{3}+3\left[\frac{9-1}{2}\right] \\ &=\frac{26}{3}+\frac{24}{2} \\ &=\frac{26+36}{3} \\ &=\frac{62}{3} \end{aligned}

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