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Use the limit de finition of the integral to compute∫_1^2x^2 dx
Related Topics
Wize University Calculus 1 Textbook > Integrals
The Definite Integral
4 Activities
Use the limit de finition of the integral to compute
∫
1
2
x
2
d
x
\displaystyle\int_1^2x^2\ dx
∫
1
2
x
2
d
x
No marks will be given on the exam if you just use integration rules when they ask you to use the limit definition of the integral
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Check Submission
More The Definite Integral Questions:
Definite Integral: Integral from Definition
Evaluate
lim
n
→
∞
∑
i
=
1
n
(
3
n
)
1
1
+
3
i
n
\displaystyle\lim_{n\rightarrow\infty}\sum_{i=1}^n\ \left(\frac{3}{n}\right)\frac{1}{\sqrt{1+\dfrac{3i}{n}}}
n
→
∞
lim
i
=
1
∑
n
(
n
3
)
1
+
n
3
i
1
Practice: Definite Integral Using Riemann Sums
Evaluate
∫
0
4
(
2
x
2
+
3
)
d
x
\displaystyle \int_{0}^{4}(2x^2 +3)\text{d}x
∫
0
4
(
2
x
2
+
3
)
d
x
using Riemann sums.
Definite Integral
Rewrite
lim
n
→
∞
∑
i
=
1
n
1
2
i
sin
(
2
i
n
)
\displaystyle \lim_{n\to\infty}\sum_{i=1}^n\frac{1}{2i}\sin\left(\frac{2i}{n}\right)
n
→
∞
lim
i
=
1
∑
n
2
i
1
sin
(
n
2
i
)
as a definite integral
Definite Integral: Integral from Definition
Evaluate
lim
n
→
∞
∑
i
=
1
n
(
3
n
)
1
1
+
3
i
n
\displaystyle\lim_{n\rightarrow\infty}\sum_{i=1}^n\ \left(\frac{3}{n}\right)\frac{1}{\sqrt{1+\dfrac{3i}{n}}}
n
→
∞
lim
i
=
1
∑
n
(
n
3
)
1
+
n
3
i
1
Definite Integral: Integral from Definition
Evaluate
lim
n
→
∞
∑
i
=
1
n
(
3
n
)
1
1
+
3
i
n
\displaystyle\lim_{n\rightarrow\infty}\sum_{i=1}^n\ \left(\frac{3}{n}\right)\frac{1}{\sqrt{1+\dfrac{3i}{n}}}
n
→
∞
lim
i
=
1
∑
n
(
n
3
)
1
+
n
3
i
1
The Definite Integral: Riemann Sums
Use Riemann sums to evaluate the definite integral
∫
2
4
x
(
x
+
2
)
d
x
\displaystyle \int_{2}^{4}x(x+2)\,\text{d}x
∫
2
4
x
(
x
+
2
)
d
x
Riemann Sums with $x^3$: Definite Integrals
Use Riemann sums to evaluate the definite integral
∫
1
3
x
2
(
x
+
3
)
d
x
\displaystyle \int_{1}^{3}x^2(x+3)\,\text{d}x
∫
1
3
x
2
(
x
+
3
)
d
x
.
Find
∫
1
2
(
x
+
2
)
d
x
\int_1^2\left(x+2\right)dx
∫
1
2
(
x
+
2
)
d
x
using Riemann sum. [No credit will be given for any other method]
Approximating Areas: Definite Integrals
The limit
lim
n
→
∞
∑
i
=
1
n
1
n
1
−
(
−
1
+
i
n
)
2
\displaystyle\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{n}\sqrt{1-\left(-1+\frac{i}{n}\right)^2}
n
→
∞
lim
i
=
1
∑
n
n
1
1
−
(
−
1
+
n
i
)
2
describes which of the following?
Riemann Sums
Evaluate
lim
n
→
∞
∑
i
=
1
n
[
5
n
[
(
1
+
5
i
n
)
2
−
3
]
]
\displaystyle \lim_{n\to\infty}\ \sum_{i=1}^n\left[\frac{5}{n}\left[\left(1+\frac{5i}{n}\right)^2-3\right]\right]
n
→
∞
lim
i
=
1
∑
n
[
n
5
[
(
1
+
n
5
i
)
2
−
3
]
]
Evaluate
∫
−
1
0
d
x
1
−
x
2
\int_{-1}^0\frac{dx}{\sqrt{1-x^2}}
∫
−
1
0
1
−
x
2
d
x
A train travels at
(
t
+
1
)
m
/
s
(t+1) \space m/s
(
t
+
1
)
m
/
s
. How far does it travel after 4 seconds?
Definite Integral
Use Riemann sums to find the area under the curve
f
(
x
)
=
2
x
−
2
f(x)=2x-2
f
(
x
)
=
2
x
−
2
on the interval
[
1
,
4
]
[1,4]
[
1
,
4
]
.
Riemann Sums: Approximation of areas under curves
If
Δ
x
=
3
n
\Delta x=\frac{3}{n}
Δ
x
=
n
3
and
x
i
=
i
Δ
x
x_i=i\Delta x
x
i
=
i
Δ
x
, then what is the integral that represents
lim
n
→
∞
Σ
i
=
1
n
x
i
(
1
+
x
i
)
2
Δ
x
\lim_{n\rightarrow\infty}\Sigma_{i=1}^nx_i\left(1+x_i\right)^2\Delta x
lim
n
→
∞
Σ
i
=
1
n
x
i
(
1
+
x
i
)
2
Δ
x
?
Practice: Definite Integral Using Riemann Sums
Q.
\textbf{Q.}
Q.
Evaluate
∫
0
4
(
2
x
2
+
3
)
d
x
\displaystyle \int_{0}^{4}(2x^2 +3)\text{d}x
∫
0
4
(
2
x
2
+
3
)
d
x
using Riemann sums.
Express this summation as an integral:
∑
i
=
1
n
π
4
π
cos
(
1
1
+
π
i
2
n
)
w
h
e
n
n
→
∞
.
\sum_{i=1}^n\frac{\pi}{4\pi}\cos\left(\frac{1}{1+\frac{\pi i}{2n}}\right)\ when\ n\ \rightarrow\ \infty.
∑
i
=
1
n
4
π
π
cos
(
1
+
2
n
π
i
1
)
w
h
e
n
n
→
∞.
Definite Integral
Rewrite
lim
n
→
∞
∑
i
=
1
n
1
2
i
sin
(
2
i
n
)
\displaystyle \lim_{n\to\infty}\sum_{i=1}^n\frac{1}{2i}\sin\left(\frac{2i}{n}\right)
n
→
∞
lim
i
=
1
∑
n
2
i
1
sin
(
n
2
i
)
as a definite integral
Practice: Definite Integral and Riemann Sum
Rewrite
lim
n
→
∞
∑
i
=
1
n
(
3
n
)
(
1
1
+
3
i
n
)
\displaystyle \lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{3}{n}\right)\left(\frac{1}{\sqrt{1+\frac{3i}{n}}}\right)
n
→
∞
lim
i
=
1
∑
n
(
n
3
)
1
+
n
3
i
1
as a definite integral.
🌶️ Evaluate the limit by recognizing the Riemann sum:
lim
n
→
∞
∑
i
=
1
n
27
i
3
n
4
.
\lim\limits_{n\rightarrow\infin}\displaystyle\sum_{i=1}^n\frac{27i^3}{n^4}.
n
→
∞
lim
i
=
1
∑
n
n
4
27
i
3
.
Enter your answer as a fraction.
Riemann Sums: Definite Integrals
Find
∫
1
2
(
x
+
2
)
d
x
\int_1^2\left(x+2\right)dx
∫
1
2
(
x
+
2
)
d
x
using Riemann sum. [No credit will be given for any other method]
Use Riemann sums to evaluate the definite integral
∫
1
3
x
2
(
x
+
3
)
d
x
\displaystyle \int_{1}^{3}x^2(x+3)\,\text{d}x
∫
1
3
x
2
(
x
+
3
)
d
x
.
Definite Integrals
Compute
lim
n
→
∞
3
n
∑
i
=
1
n
1
1
+
3
i
n
\displaystyle \lim_{n\rightarrow \infin}\frac{3}{n}\sum_{i=1}^n\frac{1}{1+\frac{3i}{n}}
n
→
∞
lim
n
3
i
=
1
∑
n
1
+
n
3
i
1
Use the limit de finition of the integral to compute
∫
1
2
x
2
d
x
\displaystyle\int_1^2x^2\ dx
∫
1
2
x
2
d
x
final114
Express this summation as an integral:
∑
i
=
1
n
π
4
π
cos
(
1
1
+
π
i
2
n
)
w
h
e
n
n
→
∞
.
\sum_{i=1}^n\frac{\pi}{4\pi}\cos\left(\frac{1}{1+\frac{\pi i}{2n}}\right)\ when\ n\ \rightarrow\ \infty.
∑
i
=
1
n
4
π
π
cos
(
1
+
2
n
π
i
1
)
w
h
e
n
n
→
∞.
[Note that the answer is not unique]
final114
Find
∫
1
2
(
x
+
2
)
d
x
\displaystyle\int_1^2\left(x+2\right)dx
∫
1
2
(
x
+
2
)
d
x
using Riemann sum. [No credit will be given for any other method]
Riemann Sums with $x^3$: Definite Integrals
Use Riemann sums to evaluate the definite integral
∫
1
3
x
2
(
x
+
3
)
d
x
\displaystyle \int_{1}^{3}x^2(x+3)\,\text{d}x
∫
1
3
x
2
(
x
+
3
)
d
x
.
The Definite Integral: Riemann Sums
Use Riemann sums to evaluate the definite integral
∫
2
4
x
(
x
+
2
)
d
x
\displaystyle \int_{2}^{4}x(x+2)\,\text{d}x
∫
2
4
x
(
x
+
2
)
d
x
Definite Integral: Integral from Definition
Evaluate
lim
n
→
∞
∑
i
=
1
n
(
3
n
)
1
1
+
3
i
n
\displaystyle\lim_{n\rightarrow\infty}\sum_{i=1}^n\ \left(\frac{3}{n}\right)\frac{1}{\sqrt{1+\dfrac{3i}{n}}}
n
→
∞
lim
i
=
1
∑
n
(
n
3
)
1
+
n
3
i
1
Evaluate
∫
−
1
0
d
x
1
−
x
2
\int_{-1}^0\frac{dx}{\sqrt{1-x^2}}
∫
−
1
0
1
−
x
2
d
x