Evaluate: $\int_{1}^{3} (x^{2} + 3 x + e^{x}) dx$

Solution

$I = \int_{1}^{3} (x^{2} + 3 x + e^{x}) dx = \int_{a}^{b} f (x) dx when f (x) = x^{2} + 3 x + e^{x}; a = 1, b = 3 h = \frac{b - a}{n} = \frac{3 - 1}{n} = \frac{2}{n} f (a + rh) = f (1 + rh) = (1 + rh)^{2} + 3 (1 + rh) + e^{1 + rh} = 4 + 5 rh + r^{2} h^{2} + e x e^{rh} = r^{2} h^{2} + 5 rh + 4 + e x e^{rh}$

$\int_{a}^{b} f (x) dx = \lim_{n \to \infty} \sum_{r = 1}^{n} hf (a + rh) \lim_{n \to \infty} \sum_{r = 1}^{n} h (r^{2} h^{2} + 5 rh + 4 + ex e^{rh}) \lim_{n \to \infty} ({\sum^{2}}_{r = 1}^{n} r^{2} h^{2} + \sum_{r = 1}^{n} {rh}^{2} + \sum_{r = 1}^{n} 4 h + e \sum_{r = 1}^{n} e^{rh} x h) \lim_{n \to \infty} (\frac{8}{n^{3}} x \frac{n (n + 1) (2 n + 1)}{6} x 5 x \frac{4}{n^{2}} x \frac{n (n + 1)}{2} + 4 x \frac{2}{n} x n + (e^{h} (\frac{e^{nh} - 1}{e^{h} - 1})) . h)$

$\int_{a}^{b} f (x) d x = \frac{4}{3} x 1 x \underset{n \to \infty}{l i m} (1 + \frac{1}{n}) x \underset{n \to \infty}{l i m} (2 + \frac{1}{n}) + \underset{n \to \infty}{l i m} 10 x 1 x (1 + \frac{1}{n}) + 8 + \underset{h \to 0}{l i m} \frac{e^{h + 1} (e^{2} - 1)}{(\frac{e^{h} - 1}{h})} = \frac{4}{3} x 1 x 2 + 10 x 1 x 1 + 8 + \frac{e (e^{2} - 1)}{1} = \frac{8}{3} + 10 + 8 + e^{3} - e = \frac{8}{3} + 18 + e^{3} - e = \frac{62}{3} + e^{3} - e$

For Daily Free Study Material Join wiredfaculty

Integrals

Evaluate: $\int_{1}^{3} (x^{2} + 3 x + e^{x}) dx$

Some More Questions From Integrals Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Integrals

Evaluate:∫13(x2 + 3x + ex) dx

Some More Questions From Integrals Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Evaluate: $\int_{1}^{3} (x^{2} + 3 x + e^{x}) dx$