Chapter 4: Q. 52 (page 353)

Calculate the exact value of each definite integral in Exercises 47–52 by using properties of definite integrals and the formulas in Theorem 4.13.
$\int_{6}^{1} (3 (1 - 2 x)^{2} + 4 x) d x$

Short Answer

Expert verified

The exact value of definite integral is $- 735$ .

Step by step solution

Step 1. Given Information

We are given,

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

Step 2. Finding the Integral

The definite integral is given by,

$\int_{6}^{1} (3 (1 - 2 x)^{2} + 4 x) d x = \int_{6}^{1} (3 (1 - 4 x + 4 x^{2}) + 4 x) d x = \int_{6}^{1} (3 - 12 x + 12 x^{2} + 4 x) d x = 3 \int_{6}^{1} 1 d x - 8 \int_{6}^{1} x d x + 12 \int_{6}^{1} x^{2} d x = 3 [1 - 6] - 8 [\frac{1}{2} (1 - 36)] + 12 [\frac{1}{3} (1 - 216)] = - 15 - 4 (- 35) + 4 (- 215) = - 15 + 140 - 860 = - 735$

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Calculate the exact value of each definite integral in Exercises 47–52 by using properties of definite integrals and the formulas in Theorem 4.13.
$\int_{6}^{1} (3 (1 - 2 x)^{2} + 4 x) d x$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the Integral

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Calculate the exact value of each definite integral in Exercises 47–52 by using properties of definite integrals and the formulas in Theorem 4.13.∫613(1-2x)2+4xdx

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the Integral

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Pure Maths

Geometry

Calculus

Theoretical and Mathematical Physics

Statistics

Study anywhere. Anytime. Across all devices.

Calculate the exact value of each definite integral in Exercises 47–52 by using properties of definite integrals and the formulas in Theorem 4.13.
$\int_{6}^{1} (3 (1 - 2 x)^{2} + 4 x) d x$