Evaluate $\int_{1}^{3} 2 x^{2} + 5 x dx$ as a limit of sum.

Solution

$\int_{1}^{3} 2 x^{2} + 5 x dx Here, a = 1, b = 3, f (x) = 2 x^{2} + 5 x ∴ n h = b - a = 3 - 1 = 2 Now \int_{a}^{b} f (x) dx = \lim_{h \to 0} h f (a) + f (a + h) + f (a + 2 h) + . . . . . + F (a + (n - 1) h) ∴ \int_{1}^{3} 2 x^{2} + 5 x dx = \lim_{h \to 0} h f (1) + f (1 + h) + f (1 + 2 h) + . . . . . + F (1 + (n - 1) h)$

$= \lim_{h \to 0} h [2 (1)^{2} + 5 (1) + 2 (1 + h)^{2} + 5 (1 + h) + (2 (1 + 2 h)^{2} + 5 (1 + 2 h)) . . . . . . . + (2 (1 + (n - 1) h)^{2} + 5 (1 + (n - 1) h))] = \lim_{h \to 0} h [7 + (2 h^{2} 9 h + 7) + (8 h^{2} + 18 h + 7) + . . . . . . . . . . + {(2 (n - 1)}^{2} h^{2} + 9 (n - 1) h + 7)] = \lim_{h \to 0} h [7 n + 2 h^{2} (1^{2} + 2^{2} + . . . . . . . . + (n - 1)^{2}) + 9 h (1 + 2 + . . . . . . . . + (n - 1)] = \lim_{h \to 0} h [7 n + 2 h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 9 h \frac{n (n - 1)}{2}]$

$= \lim_{h \to 0} h [7 n h + 2 \frac{n h (n h - h) (2 n h - h)}{6} + 9 \frac{n h (n h - h)}{2}] = \lim_{h \to 0} h [14 + 2 \frac{2 (2 - h) (4 - h)}{6} + 9 \frac{2 (2 - h)}{2}] = 14 + \frac{16}{3} + 18 = \frac{112}{3}$

For Daily Free Study Material Join wiredfaculty

Vector Algebra

Evaluate $\int_{1}^{3} 2 x^{2} + 5 x dx$ as a limit of sum.

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Vector Algebra

Evaluate ∫13 2 x2 + 5 x dx as a limit of sum.

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Evaluate $\int_{1}^{3} 2 x^{2} + 5 x dx$ as a limit of sum.