Find \( \displaystyle\int (3x^2 - 2x + 1)\, dx \).
\( x^3 - x^2 + x + C \).
Integration
12 exam-style questions on indefinite and definite integrals, areas under curves. Tap Show answer for full working.
4 Easy4 Medium4 Hard
Evaluate \( \displaystyle\int_0^2 2x\, dx \).
\( [x^2]_0^2 = 4 \).
Find \( \displaystyle\int \dfrac{1}{x^2}\, dx \).
\( \displaystyle\int x^{-2}\, dx = -x^{-1} + C = -\dfrac{1}{x} + C \).
Find the area under \( y = x \) from \( x=0 \) to \( x=4 \).
\( \displaystyle\int_0^4 x\, dx = \left[\dfrac{x^2}{2}\right]_0^4 = 8 \).
Find \( \displaystyle\int (2x+3)^4\, dx \).
Let \( u=2x+3 \), \( du=2\,dx \). \( \dfrac{1}{2}\int u^4\, du = \dfrac{1}{10}u^5 + C = \dfrac{1}{10}(2x+3)^5 + C \).
Evaluate \( \displaystyle\int_1^4 \sqrt{x}\, dx \).
\( \displaystyle\int_1^4 x^{1/2}\, dx = \left[\dfrac{2}{3}x^{3/2}\right]_1^4 = \dfrac{2}{3}(8) - \dfrac{2}{3}(1) = \dfrac{14}{3} \).
Find the area bounded by the curve \( y = x^2 \), the \( x \)-axis and the lines \( x=1 \) and \( x=3 \).
\( \displaystyle\int_1^3 x^2\, dx = \left[\dfrac{x^3}{3}\right]_1^3 = 9 - \dfrac{1}{3} = \dfrac{26}{3} \).
Given \( \dfrac{dy}{dx} = 6x^2 \) and \( y=5 \) when \( x=1 \), find \( y \) in terms of \( x \).
\( y = 2x^3 + C \). \( 5 = 2 + C \) ⇒ \( C=3 \). \( y = 2x^3 + 3 \).
Find the area enclosed between the curve \( y = 4 - x^2 \) and the \( x \)-axis.
Roots \( x=\pm 2 \). Area \( = \displaystyle\int_{-2}^{2} (4-x^2)\, dx = 2\int_0^2 (4-x^2)\, dx = 2\left[4x - \dfrac{x^3}{3}\right]_0^2 = 2\left(8 - \dfrac{8}{3}\right) = 2\cdot\dfrac{16}{3} = \dfrac{32}{3} \).
Find the area bounded by \( y = x^2 \) and \( y = 2x \).
Intersect at \( x=0,2 \). Area \( = \displaystyle\int_0^2 (2x - x^2)\, dx = \left[x^2 - \dfrac{x^3}{3}\right]_0^2 = 4 - \dfrac{8}{3} = \dfrac{4}{3} \).
Evaluate \( \displaystyle\int_0^{\pi/2} \sin x\, dx \).
\( \left[-\cos x\right]_0^{\pi/2} = 0 - (-1) = 1 \).
The curve \( y = f(x) \) passes through \( (0,2) \) and \( \dfrac{dy}{dx} = 3x^2 - 4x + 1 \). Find the equation of the curve and the area under the curve from \( x=0 \) to \( x=2 \).
\( y = x^3 - 2x^2 + x + C \). At \( (0,2) \): \( C=2 \). \( y = x^3 - 2x^2 + x + 2 \). Area \( = \displaystyle\int_0^2 (x^3-2x^2+x+2)\, dx = \left[\dfrac{x^4}{4} - \dfrac{2x^3}{3} + \dfrac{x^2}{2} + 2x\right]_0^2 = 4 - \dfrac{16}{3} + 2 + 4 = 10 - \dfrac{16}{3} = \dfrac{14}{3} \).