I = $\int\limits_{1}^{3}(3x^2+2x)$ dx Here a = 1 , b = 3 f (x) = $3x^2$ + 2x h = $\frac{b-a}{n}$ = $\frac{2}{n}$ Since, $\int\limits_{a}^{b}$ f (x) dx = $\lim\limits_{h\to 0}$ h [f (a) + f (a + h) + ... + f (a + (n - 1) h)] So, $\int\limits_{1}^{3}(3x^2+2x)$ = $\lim\limits_{h\to 0}$ h [3 $(1)^2$ + 2 (1)) + (3 $(1+h)^2$ + 2 (1 + h) + 3 $(1+2h)^2$ + 2 (1 + 2h)) ... + 3 $(1+(n-1)h)^2$ + 2 (1 + (n - 1) h)] = $\lim\limits_{h\to 0}$ h [3 (n) + 3 $(h^2+4h^2+\dots+(n-1)^2h^2)$ + 3 (2h + 4h + ... + 2 (n - 1) h) + 2n + 2 (h + 2h + ... + (n - 1) h)] = $\lim\limits_{h\to 0}$ [5nh + $3h^3(1^2+2^2+\dots+(n-1)^2)$ + $6h^2$ (1 + 2 + ... + (n - 1)) + $2h^2$ (1 + 2 + (n - 1))] = $\lim\limits_{h\to 0}$ = $\lim\limits_{h\to 0}$ = $\left[ 10 + \frac{2 \times 2 \times 4}{2} + 4 \times 2 \times 2 \right]$ = 10 + 8 + 16 = 34 OR ⇒ $\frac{x^2}{9} + \frac{y^2}{4}$ = 1 ⇒ y = $\frac{2}{3} \sqrt{9-x^2}$ Given line $\frac{x}{3} + \frac{y}{2}$ = 1 ⇒ y = $\left(2 - \frac{2x}{3}\right)$ Required Area $\{(x,y) : \frac{x^2}{9} + \frac{y^2}{4} \leq 1 \leq \frac{x}{3} + \frac{y}{2}\}$ is given below Required Area = $\int\limits_{0}^{3}(y_1 - y_2)$ dx = $\int\limits_{0}^{3}\left[ \frac{2}{3} \sqrt{9-x^2} - \left(2 - \frac{2x}{3}\right) \right]$ dx = = $\left[ \frac{2}{3} \left( \frac{9}{2} \sin^{-1} 1 \right) - 6 + 3 \right]$ - 0 = 3 × $\frac{\pi}{2}$ - 3 = $\frac{3}{2}$ (π - 2) sq units