Evaluate the definite integrals- -124 x3-5 x2-6 x-9 d x

Let I =∫_1^2(4x^3 5x^2+6x+9)dx = [ 4x^4/4 5x^3/3+6x^2/2+9x ]^2_1 = [ x^4 5x^3/3+3x^2+9x ]^2_1 = [ (2)^4 5(2)^3/3+3(2)^2+9(2) ] [ (1)^4 5(1)^3/3+3(1)^2+9(1 ...

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Byju's Answer

Standard XII

Mathematics

Theorems on Integration

Evaluate the ...

Question

Evaluate the definite integrals.
$\int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) d x$

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Solution

Let $I = \int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) d x$
$= {[\frac{4 x^{4}}{4} - \frac{5 x^{3}}{3} + \frac{6 x^{2}}{2} + 9 x]}_{1}^{2} = {[x^{4} - \frac{5 x^{3}}{3} + 3 x^{2} + 9 x]}_{1}^{4} = [(2)^{4} - \frac{5 (2)^{3}}{3} + 3 (2)^{2} + 9 (2)] - [(1)^{4} - \frac{5 (1)^{3}}{3} + 3 (1)^{2} + 9 (1)] = [16 - \frac{40}{3} + 12 + 18] - [1 - \frac{5}{3} + 3 + 9] = [46 - \frac{40}{3}] - [13 - \frac{5}{3}] = [\frac{138 - 40}{3}] - [\frac{39 - 5}{3}] = \frac{98}{3} - \frac{34}{3} = \frac{64}{3}$

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Similar questions

Q. Evaluate the definite integral

\int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) d x

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\int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) d x

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\int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) d x

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Evaluate the definite integrals. ∫21(4x3−5x2+6x+9)dx

Let I=∫21(4x3−5x2+6x+9)dx =[4x44−5x33+6x22+9x]21=[x4−5x33+3x2+9x]21=[(2)4−5(2)33+3(2)2+9(2)]−[(1)4−5(1)33+3(1)2+9(1)]=[16−403+12+18]−[1−53+3+9]=[46−403]−[13−53]=[138−403]−[39−53]=983−343=643

Evaluate the definite integrals.
$\int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) d x$