JEE Main Integration Practice Test 2026 - Free Integration Practice Questions and Study Guide

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Evaluate the integral ∫ (2x^3 - 3x^2 + x) dx.

(1/2)x^4 - x^3 + (1/2)x^2 + C

To evaluate the integral ∫ (2x^3 - 3x^2 + x) dx, we apply the power rule of integration, which states that the integral of x^n is (1/(n+1))x^(n+1) + C, for any real number n not equal to -1.

In this case, we break down the integral into the sum of the integrals of each term:

1. For the first term, 2x^3:

∫ 2x^3 dx = 2 * (1/(3 + 1))x^(3 + 1) = 2 * (1/4)x^4 = (1/2)x^4.

2. For the second term, -3x^2:

∫ -3x^2 dx = -3 * (1/(2 + 1))x^(2 + 1) = -3 * (1/3)x^3 = -x^3.

3. For the third term, x:

∫ x dx = (1/(1 + 1))x^(1 + 1) = (1/2)x^2.

Putting it all together

Get further explanation with Examzify DeepDiveBeta

(2/5)x^5 - x^3 + C

(1/3)x^3 - (3/2)x^2 + C

(1/4)x^4 - (3/3)x^2 + C

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