Definite Integrals

Definite integrals are used to find the accumulated value of a function over an interval. They often represent area under a curve, total change, displacement, or other quantities built up over time. To evaluate a definite integral, find an antiderivative and then subtract the value at the lower bound from the value at the upper bound.

Lesson

Practice Problems & Videos

\(\textbf{1)}\) \(\displaystyle\int_{2}^{6}5 \, dx\)
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The answer is \(20\)
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\(\,\,\,\,\,\,\displaystyle\int_{2}^{6}5 \, dx\)

\(\,\,\,\,\,\,\displaystyle 5x \Big|_{2}^{6}\)

\(\,\,\,\,\,\,\ 5(6)-5(2)\)

\(\,\,\,\,\,\,\ 30-10\)

\(\,\,\,\,\,\,\)The answer is \(20\)

\(\textbf{2)}\) \(\displaystyle\int_{-3}^{3}x \, dx\)
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The answer is \(0\)
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\(\,\,\,\,\,\,\displaystyle\int_{-3}^{3}x \, dx\)

\(\,\,\,\,\,\,\displaystyle \frac{x^2}{2} \Big|_{-3}^{3}\)

\(\,\,\,\,\,\,\ \frac{(3)^2}{2}-\frac{(-3)^2}{2}\)

\(\,\,\,\,\,\,\ \frac{9}{2}-\frac{9}{2}\)

\(\,\,\,\,\,\,\)The answer is \(0\)

\(\textbf{3)}\) \(\displaystyle\int_{5}^{7}x^3 \, dx\)
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The answer is \(444\)
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\(\,\,\,\,\,\,\displaystyle\int_{5}^{7}x^3 \, dx\)

\(\,\,\,\,\,\,\displaystyle \frac{x^4}{4} \Big|_{5}^{7}\)

\(\,\,\,\,\,\,\ \frac{(7)^4}{4}-\frac{(5)^4}{4}\)

\(\,\,\,\,\,\,\ \frac{2401}{4}-\frac{625}{4}\)

\(\,\,\,\,\,\,\ \frac{1776}{4}\)

\(\,\,\,\,\,\,\)The answer is \(444\)

\(\textbf{4)}\) \(\displaystyle\int_{1}^{2}\left(5x^2+4x\right) \, dx\)
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The answer is \(\frac{53}{3}\)
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\(\,\,\,\displaystyle\int_{1}^{2}\left(5x^2+4x\right) \, dx\)

\(\,\,\,\displaystyle \frac{5x^{2+1}}{2+1}+\frac{4x^{1+1}}{1+1} \Big|_1^2 \)

\(\,\,\,\displaystyle \frac{5x^{3}}{3}+\frac{4x^{2}}{2} \Big|_1^2 \)

\(\,\,\,\displaystyle \left(\frac{5(2)^{3}}{3}+\frac{4(2)^{2}}{2}\right)-\left(\frac{5(1)^{3}}{3}+\frac{4(1)^{2}}{2}\right)\)

\(\,\,\,\displaystyle \frac{53}{3}\)

\(\,\,\,\,\,\,\)The answer is \(\frac{53}{3}\)

\(\textbf{5)}\) \(\displaystyle\int_{0}^{1}\left(2x^5-3x^2\right) \, dx\)
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The answer is \(-\frac{2}{3}\)
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\(\,\,\,\displaystyle\int_{0}^{1}\left(2x^5-3x^2\right) \, dx\)

\(\,\,\,\displaystyle \frac{2x^{5+1}}{5+1}-\frac{3x^{2+1}}{2+1} \Big|_0^1\)

\(\,\,\,\displaystyle \frac{2x^{6}}{6}-\frac{3x^{3}}{3} \Big|_0^1\)

\(\,\,\,\displaystyle \left(\frac{2(1)^{6}}{6}-\frac{3(1)^{3}}{3}\right)-\left(\frac{2(0)^{6}}{6}-\frac{3(0)^{3}}{3}\right)\)

\(\,\,\,\displaystyle \frac{2}{6}-1\)

\(\,\,\,\displaystyle -\frac{2}{3}\)

\(\,\,\,\,\,\,\)The answer is \(-\frac{2}{3}\)

\(\textbf{6)}\) \(\displaystyle\int_{1}^{2}10x^9 \, dx\)
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The answer is \(1{,}023\)
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\(\,\,\,\displaystyle \int_{1}^{2}10x^9 \, dx\)

\(\,\,\,\displaystyle \frac{10x^{9+1}}{9+1} \Big|_1^2 \)

\(\,\,\,\displaystyle \frac{10x^{10}}{10} \Big|_1^2 \)

\(\,\,\,\displaystyle x^{10} \Big|_1^2 \)

\(\,\,\,\displaystyle 2^{10}-1^{10} \)

\(\,\,\,\displaystyle 1{,}024-1 \)

\(\,\,\,\,\,\,\)The answer is \(1{,}023\)

\(\textbf{7)}\) \(\displaystyle\int_{1}^{3}2x^3+3x^2 \, dx\)
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The answer is \(66\)
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\(\,\,\,\displaystyle\int_{1}^{3}2x^3+3x^2 \, dx\)

\(\,\,\,\displaystyle \frac{2x^{3+1}}{3+1}+\frac{3x^{2+1}}{2+1} \Big|_1^3 \)

\(\,\,\,\displaystyle \frac{x^{4}}{2}+x^{3} \Big|_1^3 \)

\(\,\,\,\displaystyle \left(\frac{(3)^{4}}{2}+(3)^3\right)-\left(\frac{(1)^{4}}{2}+(1)^3\right)\)

\(\,\,\,\displaystyle \left(\frac{81}{2}+27\right)-\left(\frac{1}{2}+1\right)\)

\(\,\,\,\displaystyle \frac{135}{2}-\frac{3}{2}\)

\(\,\,\,\,\,\,\)The answer is \(66\)

\(\textbf{8)}\) \(\displaystyle\int_{0}^{\pi} \, dx\)
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The answer is \(\pi\)
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\(\,\,\,\,\,\,\displaystyle\int_{0}^{\pi} 1 \, dx\)

\(\,\,\,\,\,\,\displaystyle x \Big|_{0}^{\pi}\)

\(\,\,\,\,\,\,\ (\pi) – (0)\)

\(\,\,\,\,\,\,\)The answer is \(\pi\)

\(\textbf{9)}\) \(\displaystyle\int_{1}^{2}5x^4 \, dx\)
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The answer is \(31\)
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\(\,\,\,\displaystyle \int_{1}^{2}5x^4 \, dx\)

\(\,\,\,\displaystyle \frac{5x^{4+1}}{4+1} \Big|_1^2\)

\(\,\,\,\displaystyle x^5 \Big|_1^2\)

\(\,\,\,\displaystyle \left(2^5 – 1^5 \right)\)

\(\,\,\,\displaystyle (32 – 1)\)

\(\,\,\,\,\,\,\,\)The answer is \(31\)

\(\textbf{10)}\) \(\displaystyle\int_{0}^{1} 3x^2+4 \, dx\)
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The answer is \(5\)
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\(\,\,\,\displaystyle \int_{0}^{1} 3x^2 + 4 \, dx\)

\(\,\,\,\displaystyle \frac{3x^{2+1}}{2+1} + 4x \Big|_0^1\)

\(\,\,\,\displaystyle x^3 + 4x \Big|_0^1\)

\(\,\,\,\displaystyle \left(1^3 + 4(1)\right) – \left(0^3+4(0)\right)\)

\(\,\,\,\displaystyle \left(5\right) – \left(0\right)\)

\(\,\,\,\,\,\,\)The answer is \(5\)

\(\textbf{11)}\) \(\displaystyle\int_{0}^{\pi/2} \sin(x) \, dx\)
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The answer is \(1\)
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\(\,\,\,\displaystyle \int_{0}^{\pi/2} \sin(x) \, dx\)

\(\,\,\,\displaystyle -\cos(x) \Big|_0^{\pi/2}\)

\(\,\,\,\displaystyle \left(-\cos\left(\frac{\pi}{2}\right)\right) – (-\cos(0))\)

\(\,\,\,\displaystyle (0) – (-1)\)

\(\,\,\,\,\,\,\)The answer is \(1\)

\(\textbf{12)}\) \(\displaystyle\int_{-1}^{1} x^3 \, dx\)
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The answer is \(0\)
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\(\,\,\,\displaystyle \int_{-1}^{1} x^3 \, dx\)

\(\,\,\,\displaystyle \frac{x^{4}}{4} \Big|_{-1}^1\)

\(\,\,\,\displaystyle \left(\frac{(1)^4}{4}\right) – \left(\frac{(-1)^4}{4}\right)\)

\(\,\,\,\,\,\,\)The answer is \(0\)

\(\textbf{13)}\) \(\displaystyle\int_{0}^{2} (2x + 1) \, dx\)
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The answer is \(6\)
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\(\,\,\,\displaystyle \int_{0}^{2} (2x + 1) \, dx\)

\(\,\,\,\displaystyle x^2 + x \Big|_{0}^2\)

\(\,\,\,\displaystyle \left(2^2 + 2\right) – \left(0^2 + 0\right)\)

\(\,\,\,\,\,\,\)The answer is \(6\)

\(\textbf{14)}\) \(\displaystyle\int_{-2}^{2} 4 \, dx\)
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The answer is \(16\)
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\(\,\,\,\displaystyle \int_{-2}^{2}4 \, dx\)

\(\,\,\,\displaystyle 4x \Big|_{-2}^{2}\)

\(\,\,\,\displaystyle \left(4(2)\right) – \left(4(-2)\right)\)

\(\,\,\,\displaystyle \left(8\right) – \left(-8\right)\)

\(\,\,\,\,\,\,\)The answer is \(16\)

\(\textbf{15)}\) \(\displaystyle\int_{1}^{4} 3x \, dx\)
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The answer is \(\displaystyle \frac{45}{2}\)
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\(\,\,\,\displaystyle \int_{1}^{4}3x \, dx\)

\(\,\,\,\displaystyle \frac{3x^{2}}{2} \Big|_{1}^{4}\)

\(\,\,\,\displaystyle \frac{3(4)^2}{2} – \frac{3(1)^2}{2}\)

\(\,\,\,\displaystyle \frac{48}{2} – \frac{3}{2}\)

\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{45}{2}\)

\(\textbf{16)}\) \(\displaystyle\int_{1}^{2} (6x^2 + 5x) \, dx\)
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The answer is \(\displaystyle \frac{43}{2}\)
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\(\,\,\,\displaystyle \int_{1}^{2} (6x^2 + 5x) \, dx\)

\(\,\,\,\displaystyle 2x^{3} + \frac{5x^{2}}{2} \Big|_{1}^{2}\)

\(\,\,\,\displaystyle \left(2(2)^3 + \frac{5(2)^2}{2}\right) – \left(2(1)^3 + \frac{5(1)^2}{2}\right)\)

\(\,\,\,\displaystyle \left(16 + \frac{20}{2}\right) – \left(2 + \frac{5}{2}\right)\)

\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{43}{2}\)

\(\textbf{17)}\) \(\displaystyle\int_{-1}^{1} x^5 \, dx\)
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The answer is \(0\)
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\(\,\,\,\displaystyle \int_{-1}^{1} x^5 \, dx\)

\(\,\,\,\displaystyle \frac{x^{6}}{6} \Big|_{-1}^{1}\)

\(\,\,\,\displaystyle \left(\frac{(1)^6}{6}\right) – \left(\frac{(-1)^6}{6}\right)\)

\(\,\,\,\displaystyle \frac{1}{6} – \frac{1}{6}\)

\(\,\,\,\,\,\,\)The answer is \(0\)

\(\textbf{18)}\) \(\displaystyle\int_{0}^{4}\sqrt{x}\,dx\)
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The answer is \(\displaystyle \frac{16}{3}\)
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\(\,\,\,\displaystyle\int_{0}^{4}\sqrt{x}\,dx\)

\(\,\,\,\displaystyle\int_{0}^{4}x^{1/2}\,dx\)

\(\,\,\,\displaystyle \frac{x^{3/2}}{3/2}\Big|_0^4\)

\(\,\,\,\displaystyle \frac{2}{3}x^{3/2}\Big|_0^4\)

\(\,\,\,\displaystyle \frac{2}{3}(4)^{3/2}-\frac{2}{3}(0)^{3/2}\)

\(\,\,\,\displaystyle \frac{2}{3}(8)\)

\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{16}{3}\)

\(\textbf{19)}\) \(\displaystyle\int_{1}^{e}\frac{1}{x}\,dx\)
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The answer is \(1\)
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\(\,\,\,\displaystyle\int_{1}^{e}\frac{1}{x}\,dx\)

\(\,\,\,\displaystyle \ln|x|\Big|_1^e\)

\(\,\,\,\displaystyle \ln(e)-\ln(1)\)

\(\,\,\,\displaystyle 1-0\)

\(\,\,\,\,\,\,\)The answer is \(1\)

\(\textbf{20)}\) \(\displaystyle\int_{0}^{\pi/2}\cos(x)\,dx\)
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The answer is \(1\)
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\(\,\,\,\displaystyle\int_{0}^{\pi/2}\cos(x)\,dx\)

\(\,\,\,\displaystyle \sin(x)\Big|_0^{\pi/2}\)

\(\,\,\,\displaystyle \sin\left(\frac{\pi}{2}\right)-\sin(0)\)

\(\,\,\,\displaystyle 1-0\)

\(\,\,\,\,\,\,\)The answer is \(1\)

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